Count number of lines in a file using DOS

by Brett on February 25, 2007

Every once in a while I get on a computer and I need to count the number of lines in a file. My first instinct is to open my text editor (editplus) and hit ctrl+end to get to the bottom of the document. Then I can view the status bar which will tell me the line number. This works fine when I am on my computer but not when I am on another computer that does not have editplus installed.

My next option might be to open this file in notepad and do the exact same thing. This will work fine if the file is not to large. The problem is I often deal with very large files. I need a quicker way to produce the same results.

This is where DOS comes into play. I can use the following command and let DOS quickly tell me the number of lines in the file.

findstr /R /N "^" file.txt

This command will output every line with a line number in front of it but will still take a long time given a very large file. The solution to this is to take this command one step further.

findstr /R /N "^" file.txt | find /C ":"

Now the output will only be the number of lines that are contained in the file.Again, this command could be taken a step further to tell you how many lines are in the file that contain a certain string.

findstr /R /N "^.*certainString.*$" file.txt | find /c ":"

I’m sure there are many other great uses for find and findstr. If you have found one please post comment.

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{ 59 comments… read them below or add one }

Ken Happich June 3, 2008 at 11:36 pm

I find this to be a very interesting use of findstr and like your website. I recently ran into an issue/limitation with findstr that I did not know existed.

I was trying to find a pattern at a particular column location in the file and it might have been at column position aprox 140. I used the following command

findstr ^…………………………………………………..pattern *.txt>testOutput.txt

In my case there were serveral more periods (single charter wildcard) and I got the error message search string too long.

Is there anyway with findstr to find a pattern at a unique location in a very long line when the pattern is closer to the end of the line?

Reply

Brett June 4, 2008 at 1:17 pm

I think the below searches might help you if I understand the problem correctly.

The below line will find “pattern” anywhere in a line.
findstr /R ".*pattern.*" afile.txt

The below line will find “pattern” after 140 characters.
findstr /r "^(.*){140}pattern.*$" afile.txt

Reply

Sponge Belly May 25, 2013 at 10:22 am

Hi Brett!

No need for .* in the first example.

findstr /r “pattern” afile.txt

will do. And findstr doesn’t support the () and {} syntax used in the second example, sorry.

What Ken wants to achieve is very difficult using findstr. Maybe if he knew that the pattern always ended five characters before the end of line, he could try something like:

findstr /r “pattern…..$” afile.txt

but that outputs all lines containing a match, not merely all text that matches findstr’s regular expression.

Findstr is frustratingly limited and buggy. Ken would be better off using JavaScript. JS can be embedded inside Batch files (search for “batch javascript hybrid”) so you can have the best of both worlds.

Reply

iNFiNiTyLoOp June 15, 2008 at 10:45 pm

find /c/v "" afile.txt
Is the way I figured out how to do it.

Reply

Sponge Belly May 25, 2013 at 10:34 am

@InfinityLoop

The problem with:

find /c /v “” afile.txt

is that it outputs filename as well as number of lines:

find /c /b “” < afile.txt

gets around this by using redirection, but count can be incorrect if file ends with many blank lines.

Either method fails to deal with null characters which the find command outputs as newlines. And additional newlines would cause an incorrect count.

Reply

Rahul Babar July 2, 2009 at 8:00 pm

We can open file using dos editor which will show the line numbers you are at.
No need to remember special dos command.
Simply put dos command like

>edit file.txt

Thanks,
Rahul

Reply

Faraz January 4, 2011 at 9:47 am

Liked your line count command line.
findstr /R /N "^" file.txt | find /C ":"

I want that if I get 44 lines count from the text file then I do not want to do any thing but if it is greater or less than 44 then Exit Batch file.

Can you help me please.

Thanks

Reply

Brett January 4, 2011 at 10:56 am

@Rahul: that will not work on 64 bit versions of windows

@Faraz: You can do that in a batch file with something like the following. Just change the name of yourfile.txt and possibly the temp file (save_temp_count.txt):

@echo off
:: Get the count and save it in a temp file
findstr /R /N "^" yourfile.txt | find /C ":" > save_temp_count.txt
:: Get the contents of the temp file and save it to the %count% variable
set /p count=<save_temp_count.txt
:: If greater or less than 44 (not 44) stop batch file
if NOT %count%==44 (
	GOTO END
)
:: count is 44
echo "count is 44"
:END

Reply

Faraz January 4, 2011 at 11:24 am

Brett
Thank you for your prompt reply.
I tested the code, batch file stuck at this below line.
set /p count=

Thanks

Reply

Brett January 4, 2011 at 12:00 pm

@Faraz: Sorry, the code suffered from a bad paste. I have updated the above code.

Let me know if it works for you.

Reply

Faraz January 4, 2011 at 9:00 pm

Brett: Brilliant!!! It worked Great. Thank you very much, God bless you & your family.

Reply

Moesjamarra February 2, 2011 at 4:54 am

Works perfect! Thx a lot!

Reply

Nogs February 3, 2011 at 12:32 am

Hi,
thanks a lot for your post, I still have a question:
I want to use this line and store the count result in a variable, can anyone help?

Thanks.

Reply

Nogs February 3, 2011 at 12:42 am

sorry, I read more carefully and this post already have what I needed.
thanks it helped a lot!

Reply

Faraz March 7, 2011 at 2:38 am

Hi Brett,

This below dos code when I run in batch file it Rename all files in Numeric sequence like 1.jpg, 2.jpg, 3.jpg, 4.jpg……. one problem with this code. suppose we have 99 files, When it rename the last file 99.jpg then at the end it again rename 9.jpg to last file like 100.jpg do u have a clue to fix it so it don’t do that or have better dos batch file code to do this task.
===================================================

@echo off
set /a cnt=1
for %%a in (*.jpg) do call :PROCESS “%%a”
goto :EOF
:PROCESS
ren %1 %cnt%.jpg
set /a cnt+=1

===================================================

Thanks

Reply

Tom March 16, 2011 at 8:47 am

Here’s another example that i found useful:

type *.txt | findstr /R /N “^” | find /C “:”

good for use to count total lines of all the files.

Reply

mark March 24, 2011 at 9:17 am

awesome, helped on my uni tutorial, tears since I used dos as mature student
top stuff =)

Reply

FractalSpace April 6, 2011 at 2:33 pm

Nice though, but Google brought me here on a different search. How to display current line number in a batch file (for debugging) ?

Reply

john July 19, 2011 at 6:49 pm

Hello,

I have a batch file that counts the number of lines in each text file inside a directory, and then prepends the total number of lines in each file to the beginning of each file in the directory. It looks like this:

@echo off
setlocal enabledelayedexpansion
cd /d "C:\Users\John\Desktop\Testing\AllFiles"
for /f "delims=" %%F in ('dir /b *.txt') do (
   for /f "delims=:" %%N in ('findstr /N /R "^" "%%~dpnxF"') do set lines=%%N
      echo There are !lines! lines in this file > "%%~dpnxF.temp"
      type "%%~dpnxF">>"%%~dpnxF.temp"
      del "%%~dpnxF"
      echo "%%~dpnxF.temp" "%%~nxF"
      ren "%%~dpnxF.temp" "%%~nxF"
      )

The batch file above works great for small text files in the directory, however, it takes a long time to read the large text files in the directory.

As you stated in your blog, using the code

findstr /R /N “^” file.txt

Does not work well for counting the lines in large text files. So, i added the following code:

findstr /R /N “^” file.txt | find /C “:”

from your blog into the batch file, and now it looks like this:

@echo off
setlocal enabledelayedexpansion
cd /d "C:\Users\John\Desktop\Testing\AllFiles"
for /f "delims=" %%F in ('dir /b *.txt') do (
   for /f "delims=:" %%N in ('findstr /N /R "^" "%%~dpnxF" | find /C ":"') do set lines=%%N
      echo There are !lines! lines in this file > "%%~dpnxF.temp"
      type "%%~dpnxF">>"%%~dpnxF.temp"
      del "%%~dpnxF"
      echo "%%~dpnxF.temp" "%%~nxF"
      ren "%%~dpnxF.temp" "%%~nxF"
      )

But, when i run the modified code, i get the following error:

“: was unexpected at this time”

Do you know what i am doing wrong?

Thanks in advance :)

Reply

Dave Shibli January 25, 2012 at 6:57 am

I had a need to number a text file before bulk inserting into a database so I could keep the lines in order.
findstr /R /N “^” file.txt >> newfile.txt saved my life…Thanks much.
I wrapped it in a command shell and loaded the file names from SQL and indexed nearly 5,000 text files for a forensic analysis.

Reply

Jadh March 15, 2012 at 2:49 pm

This does not work for really big files such as 10 million row. Do you know how to make it work for big files?

Reply

Brett March 15, 2012 at 4:07 pm

@Jadh, I just tried this on a file that had 100 million lines (I added 90 million to your number). I used the following command and it worked perfectly.

findstr /R /N “^” file.txt | find /C “:”

It took a few minutes for it to go through the file but it eventually output 100,000,000. I ran this on Windows 7 64 bit with 8GB memory.

Reply

Jadh March 15, 2012 at 4:12 pm

Yup you are right. I was reconciling different records…Thanks for the quick resposne.

Reply

Robert April 10, 2012 at 1:41 pm

Using Brett’s findstr command string, I wrote the following to poll the connection status of my Android device over ADB. The script keeps looping until a device is connected.

echo OFF

:START

adb devices > adb_devices.txt
findstr /R /N “^” adb_devices.txt | find /C “:” > linecount.txt
set /p count=<linecount.txt
if %count% LEQ 2 GOTO START

:END

Reply

ss July 12, 2012 at 2:50 am

Hi Brett

I used findstr /R /N “^” test.xml
I renamed test.xml to test.xml1 then also the command worked.
I was expecting command will fail since test.xml is no longer there, but it still worked. Can you please tell me the reason.

Reply

Brett November 20, 2012 at 6:50 pm

It fails for me on Windows 7. I ran the following commands and got an error when I tried to run findstr on a file that did not exist:

c:\test>mv test.txt test.xml
c:\test>findstr /R /N "^" test.xml | find /C ":"
2
c:\test>mv test.xml test.xml1
c:\test>findstr /R /N "^" test.xml | find /C ":"
FINDSTR: Cannot open test.xml
0

Reply

abha September 8, 2012 at 9:10 am

Hi,
I want to delete the last 2 lines of the file in DOS.
I do not want to read each line and copy it to a new file. As the file contains thousands of record and it is taking too much time.
Basically i want to copy all the contents of a file except the last 2 lines.
I also have the total number of lines..Just if it makes the task easy.
Please help……

Reply

Brett November 20, 2012 at 6:45 pm

I believe you are going to need a programming language for this. I do not know of a way to do this with the Windows Command Prompt. It could probably be done in Windows Powershell though but that’s not very far from using a programming language.

With a quick Google I found a script someone put together for this using python. I haven’t tested it but you can view the script here.

Reply

Anupam November 20, 2012 at 5:54 pm

To count the number of lines in a file why would be do
findstr /R /N “^” file.txt | find /C “:”
Instead one can use
find /C /V “” file.txt
Right?
Also for counting the occurance of particular string in a file what is wrong in using find /C “StringToSearch” file.txt ?

Thanks,
Anupam

Reply

Brett November 20, 2012 at 6:27 pm

I ran a quick test using find /C /V “” file.txt where the file contained the following:

test1
 
 
test2
 
 
 

and it produced the output of:

———- TEST.TXT: 5

However, the output should have been greater. Also, the output includes more than just the number.

This is back in 2007 that I put together this command but I suspect I needed only the number. I probably used the number in a batch script for another purpose.

Reply

Anupam November 20, 2012 at 5:59 pm

Moreover findstr /R /N “^” file.txt | find /C “:” gives wrong results if there are : in a file which is very common for log files with timestamps.

Reply

Brett November 20, 2012 at 6:36 pm

Actually the command works correctly with files that have colon’s. The reason is that the first part of the command:

findstr /R /N "^" file.txt

Searches file.txt using a regex of ^. Where ^ means the start of the line.

For a file containing the following:

test:1
test:2

The above part of the command will output:

1:test:1
2:test:2

Then the second part of the command find /C “:” searches on this output which produces 2 as there are exactly two lines with colons in them.

Reply

Anupam November 21, 2012 at 12:24 am

Yes I think you are right. I will check why I was getting wrong results.

Reply

satish December 27, 2012 at 5:18 am

i have executed sql file and generated logfile. it has around 80000 lines. But it has 2 empty lines between every two lines.
i have used
findstr /R /N “^.*certainString.*$” filename.txt
but i am get wrong output. mybe because of gap between lines.
so can you give me a fix for this ? and i need to use this command frequently.

Reply

Brett December 28, 2012 at 9:53 am

@satish – It sounds like you just want to ignore the blank lines. Assuming that there are no spaces or funny characters on the blank lines you could do something like the following:

findstr /N /R “^[^\n]” c:\test.txt | find /c “:”

This will search for and count any line that does not start with a newline character of \n.

Reply

satish January 1, 2013 at 11:57 pm

Hi Brett,
Thanks so much for your quick reply. following statement has fulfilled my requirement.

findstr  "string"  filename  | find  /c":"

The blank lines are ignored when it is searching for the “string” .
thanks for your post.and Happy new year………

Reply

Partha February 25, 2013 at 6:46 am

Hey, your post is great…I need a help with a script …to find and replace a string in a xml file..preferablly using regular expression.
Your code to find the number of occurance works great,..I need script to find and replace a string…

Can you help ..

Regards
Partha

Reply

Steve March 18, 2013 at 10:12 am

This…. is awesome. Thanks very much for posting. Just used this on a 4.6m line file (thank goodness)

Reply

Maria March 22, 2013 at 1:54 pm

hi,
i’m greatful if anyody can help me with this issue. i want to get a variable in the 1st line of a txt file and store it in a new variable
aaa.txt
yymmdd 010 xxx
aaa
bbb
I want to open the file, store the seq-no which is 010 to a variable and close the file

Reply

Brett June 3, 2013 at 12:50 pm

Maria, the following commands might work for you. It makes the assumption that you only want to look at the first line and that the sequence number is always the 7th character in and 3 characters long.

:: Get the first line of the text file
set /p yourVariable=<test.txt
:: Get the substring at position 7 with a length of 3 characters
set yourVariable=%yourVariable:~7,3%
echo %yourVariable%

Sponge Belly also posted another solution that uses a for loop. I haven’t tested that solution.

Reply

Kim May 22, 2013 at 11:48 am

Brett-

I am trying to write a batch file to be run in a dos command prompt on XP. I am trying to get a listing of files in a specific path that follow a certain naming convention. I need to copy and rename each file instance to a static name and drop it to a transmission folder. Since it may take a little while for the file to go in the transmission folder, I need to check before I copy the next file over so that I don’t overlay the previous file. I am not able to use SLEEP or TIMEOUT since I don’t have the extra toolkit installed. I try to just continually loop back to a START section until the file is sent. I noticed that if I passed the %%x value set in the for loop that if I loop back to the START section a couple of times, it seems to loose its value and it is set to nothing. So I tried to set a variable to hold the value. I seem to be having issues with the variable not being set correctly or not cleared. Originally it kept on referencing the first file but now it doesn’t seem to be set at all. The ECHO displays the correct the value but the filename variable is empty still. Does anyone know a better way of doing this? Thanks in advance for your help as I have already wasted a whole day on this!

Thanks!!

Kim

    @ECHO "At the start of the loop"
    @for %%x in (C:\OUTBOUND\customer_file*) do (
    @ECHO "In the loop"
    @ECHO "loop value ="
    @ECHO %%x
    SET filename=%%x
    @ECHO "filename ="
    @ECHO %filename%
    @ECHO ...ARCHIVE OUTBOUND CUSTOMER FILE
    archivedatafile --sourcefile="%filename%" --archivefolder="..\archivedata\customer" --retentiondays=0
    IF NOT %ERRORLEVEL%==0 GOTO ERROR
    PAUSE
    :START
    IF EXIST l:\OutputFile (
	@ping 1.1.1.1 -n 1 -w 30000
	GOTO START
    ) ELSE (
	COPY %filename% l:\OutputFile /Y
	IF NOT %ERRORLEVEL%==0 GOTO ERROR
	PAUSE
    )
    )
    GOTO END
    :ERROR
    @echo off
    @ECHO *************************************************************
    @ECHO *                      !!ERROR!!                            *
    @ECHO *************************************************************
    :END
    SET filename=

Reply

Sponge Belly May 24, 2013 at 3:04 pm

Hi All!

The count will be incorrect if there are null characters in the input to the find command because find outputs them as newlines.

The program below uses only findstr (and some chicanery). It’s fast on large files, copes with extremely long lines, and doesn’t mind if the input file has Windows or Unix line endings.

@echo off & setlocal enableextensions
(set lf=^
)
call :LFcount lc "%~1"
echo(file "%~1" has %lc% lines
endlocal & goto :EOF
:LFcount
setlocal enabledelayedexpansion
findstr /mv "!lf!" "%~2" >nul && (
for /f delims^=: %%n in ('
cmd /v:on /c findstr /nv "^!lf^!" "%~2" 2^>nul
') do set lastline=%%n) || (for /f delims^=: %%n in ('
(findstr /n "^" "%~2" ^& echo(#^) ^| findstr /bn # 2^>nul
') do set /a lastline=%%n-1)
endlocal & set "%1=%lastline%" & exit /b 0

For more info, read this blog post: http://wp.me/p2x3If-4i

Reply

Sponge Belly May 24, 2013 at 3:15 pm

@Maria

Try this:

@echo off & setlocal enableextensions
for /f "usebackq tokens=2" %%s in ("aaa.txt") do (
set seqno=%%s
goto break
)
:break
echo(seq no is %seqno%
endlocal & exit /b 0

Copy & Paste the above into a file and call it seqno.cmd. Open a Command Prompt in the folder where you saved the file, and type

seqno

on its own and you should see the result. Let me know if you have any more problems.

Reply

AlSalman June 2, 2013 at 8:38 am

Hello;

Thanks all for your input, that was helpful to me…

I used this command

findstr /R /N "^" test*.txt | find /C ":"

To find the total number of lines in all test*.txt files (3 files) but got wrong count (13 Lines)!
Each file has 5 lines (no blank lines at the top, bottom or in the middle)….

Using this command shows why:

findstr /R /N "^" test*.txt

Output is:

test1.txt:1:a0
test1.txt:2:a1
test1.txt:3:a2
test1.txt:4:a3
test1.txt:5:a0test2.txt:1:b0
test2.txt:2:b1
test2.txt:3:b2
test2.txt:4:b3
test2.txt:5:b0test3.txt:1:c0
test3.txt:2:c1
test3.txt:3:c2
test3.txt:4:c3
test3.txt:5:c0

It is like each last line of a file is counted as one with the first line of the next file???

Any advise how to solve this?

Thanks

Reply

Brett June 3, 2013 at 12:24 pm

A for loop should help with this.

(for %a in (test*.txt) DO @(findstr /R /N "^" %a & echo:)) | find /c ":"

The above command will loop through every file (named test*.txt) and run findstr on each file followed by a newline (echo:). After every file has been processed by findstr then we will count the number of lines.

Reply

AlSalman June 17, 2013 at 11:09 pm

Thank Brett;

Thats worked very well…

Regards

Reply

Michael Q June 3, 2013 at 11:52 am

Thanks for this page. I think you should add multiple strings to this as well.

I did the following to look for XX and YY at the same time:

findstr /c:"XX" /c:":YY" /n file.txt | find /c ":" > temp_linecount.txt
set /p TCOUNT=<temp_linecount.txt
if exist temp_linecount (del temp_linecount)
echo %TCOUNT%

Reply

Satish July 25, 2013 at 5:40 am

Hi all, I have a query.
i have a text file with the contents as :
0000003834,0000002076,04/16/2013,0000000030,0000002367,0000000005,
0000003834,0000002076,04/16/2013,0000000030,0000002366,0000000000
0000003834,0000002076,04/16/2013,0000000031,0000007714,0000002366,
0000003834,0000002076,04/16/2013,0000000031,0000007716,0000002367,

These are comma separated.
Please help me in writing a batch file that searches for “0000000030″ on the 4th column and if it finds “000000000″ to the corresponding 5th column then delete the entire row

Reply

Gaurav July 26, 2013 at 4:09 am

Hi Brett,

I have below requirement. Below are two files File 1 and File 2.
File1:
abc hello
abc hi
abc we
abc changes start
how are you?
I am fine.
What about you?
I am fine.
abc changes end
File 2:
xyz hi
xyz hello
xyz yup
xyz changes start
Whatsup?
Nothing
Okay
xyz changes ends

In File 1 “abc” is a string here I want to find the count of string “abc” and then I have to add that count to the lines between abc changes start and abc changes end strings.
In same way I have to find for File2 which has “xyz” string and in the end I have to add the total count.
For example: File 1 will have a total count of “abc” string + count of lines between “abc changes start” and “abc changes ends” that is 3+4 =7 and File 2 will be having for xyz and changes start and changes end string 3+3 =6 and in the end total count will be 7+6=13.

Thanks in advance.

Reply

Vishnu July 31, 2013 at 4:39 am

Brett,

I have to go to last line of a log file and check a string , string will be available on many lines. I have to particularly check the string in last line . Please provide your concerns.

-Vishnu.

Reply

amit August 26, 2013 at 11:16 am

can we get a count of characters in a text file and then compare it?

Reply

Denis Pavlyukov September 25, 2013 at 12:52 am

Thank you, Brett, for taking your time posting what you’ve revealed. “The country will not forget you” (c)

Reply

Amit September 26, 2013 at 10:37 am

Thanks. very helpful website. I need help writing a DOS command to do “record count” in a file and ignore if any “new line” character in the file.
file has | delimeter.
Code I am using to find count:

set filea=abc.txt
set /a cnt=0
for /f %%a in (‘type “%file%”^|find “” /v /c’) do set /a cnt=%%a+1
echo %file% has %cnt% lines

For Ex:
L1234|John D Pinto| abc
I will get 1 count. Thats correct.

Now if my data will be like:
L1234|John D
Pinto| abc

In 2 lines then also I want to count as 1 record however parameter is counting as 2 count as it is currently doing line count NOT record count. Any help will be appriciated.
Thanks for reading.

Reply

Frank Jeppesen October 14, 2013 at 11:47 pm

Saved my day, trying to transport a perl program from running on unix with system commands using ‘wc -l’.
My windows rewrite was less than pretty .. and here I saw that it cound be done in one line !! 8))
Cheers from Copenhagen, Frank

Reply

saravanan CM October 30, 2013 at 5:28 am

The command => findstr /R /N “^” file.txt is very nice to see the line.But it seems takes long time to display all the records .Is possible to display last line with line number instead of displaying all the line numbers

Reply

Ian November 7, 2013 at 12:00 pm

Any way I can modify your script that you posted on May 24, 2013 at 3:04 pm
to output the last line of the file after the count has been output ?

I’ve tried for loop and skipping to the end – 1 but nothing is returned. I’m new to .bat file syntax so maybe I’m putting it in the wrong place. Any guidance much appreciated

Reply

Anirban November 11, 2013 at 12:51 am

hi brett..i have a requirement to count the number of lines in a file, but the count should exclude commented lines (i.e. lines which have * in position 1 or 7). actually, i was dealing with cobol programs, which is why the commented lines would appear in this position. can this be done using dos scripting?
thanks
anirban

Reply

Manoj December 6, 2013 at 1:51 am

Hi,

I have a text file containing 5 million records, i need to remove the trailing spaces for each record.
Could you help me please. I have a script which does this but it woks well only smaller files say 250 MB.
When input file is 2 GB, output file is returned as empty ? Why so ?

@echo off
setlocal enableDelayedExpansion
set MYDIR=C:\Documents and Settings\test
set “spcs= ”
for /l %%n in (1 1 12) do set “spcs=!spcs!!spcs!”
findstr /n “^” “%~1″>”%~1.tmp”
setlocal disableDelayedExpansion
(
for /f “usebackq delims=” %%L in (“%~1.tmp”) do (
set “ln=%%L”
setlocal enableDelayedExpansion
set “ln=!ln:*:=!”
set /a “n=4096″
for /l %%i in (1 1 13) do (
if defined ln for %%n in (!n!) do (
if “!ln:~-%%n!”==”!spcs:~-%%n!” set “ln=!ln:~0,-%%n!”
set /a “n/=2″
)
)
echo(!ln!
endlocal
)
) >”%~1″
del “%~1.tmp” 2>nul

move “%~1″ “%MYDIR%”
echo. 2>”%MYDIR%file_is_ready.txt”

Reply

Anna February 22, 2014 at 10:51 am

I need to Count the No : Of Records in the File without the Header Count.
For ex: If the File Contains 20 Rows with the Header , I need the Count to be displayed as 19.

The Below Command works fine , but the Row Count should be Minus the Header (-1)

findstr /R /N “^” file.txt | find /C “:”

Thanks!

Reply

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