How to replace one char to get many strings in Shell?

up vote
5
down vote

favorite

I can find many questions with answers in the other direction, but unfortunately not in the one I would like to have my replacements: I intend to replace a char, such as #, in a string, such as test#asdf, with a sequence, such as 0..10 to get a sequence of strings, in this example test0asdf test1asdf test2asdf test3asdf test4asdf test5asdf test6asdf test7asdf test8asdf test9asdf test10asdf.

I tried, from others:

echo '_#.test' | tr # 0..10 (throws usage)

echo '_#.test' | sed -r 's/#/0..10/g' (returns _0..10.test)

echo '_#.test' | sed -r 's/#/'0..10'/g' (works for the first, afterwards I get sed: can't read (...) no such file or directory)

What is a working approach to this problem?

Edit, as I may not comment yet: I have to use # in the string, in which this character should be replaced, as the string passed from another program. I could first replace it with another char though.

edited Aug 27 at 18:31

asked Aug 27 at 18:18

BernhardWebstudio

284

Why not just echo _0..10.test? The way you have it, it's echoing the string first and then replacing the character which isn't going to work. You could also replace the character and then echo.
â€“Â Nasir Riley
Aug 27 at 19:02

add a commentÂ |Â

up vote
5
down vote

favorite

I tried, from others:

echo '_#.test' | tr # 0..10 (throws usage)

echo '_#.test' | sed -r 's/#/0..10/g' (returns _0..10.test)

echo '_#.test' | sed -r 's/#/'0..10'/g' (works for the first, afterwards I get sed: can't read (...) no such file or directory)

What is a working approach to this problem?

edited Aug 27 at 18:31

asked Aug 27 at 18:18

BernhardWebstudio

284

Why not just echo _0..10.test? The way you have it, it's echoing the string first and then replacing the character which isn't going to work. You could also replace the character and then echo.
â€“Â Nasir Riley
Aug 27 at 19:02

add a commentÂ |Â

up vote
5
down vote

favorite

I tried, from others:

echo '_#.test' | tr # 0..10 (throws usage)

echo '_#.test' | sed -r 's/#/0..10/g' (returns _0..10.test)

echo '_#.test' | sed -r 's/#/'0..10'/g' (works for the first, afterwards I get sed: can't read (...) no such file or directory)

What is a working approach to this problem?

edited Aug 27 at 18:31

asked Aug 27 at 18:18

BernhardWebstudio

284

I tried, from others:

echo '_#.test' | tr # 0..10 (throws usage)

echo '_#.test' | sed -r 's/#/0..10/g' (returns _0..10.test)

echo '_#.test' | sed -r 's/#/'0..10'/g' (works for the first, afterwards I get sed: can't read (...) no such file or directory)

What is a working approach to this problem?

edited Aug 27 at 18:31

asked Aug 27 at 18:18

BernhardWebstudio

284

edited Aug 27 at 18:31

asked Aug 27 at 18:18

BernhardWebstudio

284

asked Aug 27 at 18:18

BernhardWebstudio

284

asked Aug 27 at 18:18

BernhardWebstudio

284

Why not just echo _0..10.test? The way you have it, it's echoing the string first and then replacing the character which isn't going to work. You could also replace the character and then echo.
â€“Â Nasir Riley
Aug 27 at 19:02

add a commentÂ |Â

Why not just echo _0..10.test? The way you have it, it's echoing the string first and then replacing the character which isn't going to work. You could also replace the character and then echo.
â€“Â Nasir Riley
Aug 27 at 19:02

Why not just echo _0..10.test? The way you have it, it's echoing the string first and then replacing the character which isn't going to work. You could also replace the character and then echo.
â€“Â Nasir Riley
Aug 27 at 19:02

add a commentÂ |Â

6 Answers
6

active

oldest

votes

up vote
6
down vote

accepted

The 0..10 zsh operator (now also supported by a few other shells including bash) is just another form of csh-style brace expansion.

It is expanded by the shell before calling the command. The command doesn't see those 0..10.

With tr '#' 0..10 (quoting that # as otherwise it's parsed by the shell as the start of a comment), tr ends being called with ("tr", "#", "0", "1", ..., "10") as arguments and tr doesn't expect that many arguments.

Here, you'd want:

echo '_'0..10'.test'

for echo to be passed "_0.test", "_1.test", ..., "_10.test" as arguments.

Or if you wanted that # to be translated into that 0..10 operator, transform it into shell code to be evaluated:

eval "$(echo 'echo _#.test' | sed 's/#/0..10/')"

where eval is being passed echo _0..10.test as arguments.

(not that I would recommend doing anything like that).

edited Aug 27 at 19:24

answered Aug 27 at 18:32

StÃ©phane Chazelas

283k53521855

add a commentÂ |Â

up vote
6
down vote

You can split the string on the delimiter, capture the prefix and the suffix, then use brace expansion to generate the names:

str='test#asdf'
IFS='#' read -r prefix suffix <<<"$str"
names=( "$prefix"0..10"$suffix" )
declare -p names

declare -a names='([0]="test0asdf" [1]="test1asdf" [2]="test2asdf" [3]="test3asdf" [4]="test4asdf" [5]="test5asdf" [6]="test6asdf" [7]="test7asdf" [8]="test8asdf" [9]="test9asdf" [10]="test10asdf")'

edited Aug 27 at 20:42

answered Aug 27 at 19:07

glenn jackman

47k265103

add a commentÂ |Â

up vote
2
down vote

Do you have to use #? Maybe you could use %d?

$ for i in 1..10; do printf "_%d.test " "$i"; done
_1.test _2.test _3.test _4.test _5.test _6.test _7.test _8.test _9.test _10.test

answered Aug 27 at 18:29

DopeGhoti

40.7k54979

add a commentÂ |Â

up vote
2
down vote

First # starts a comment. So, you need to escape it with .

Second, use a for loop.

Here is your solution:

for i in 1..10
do
 echo '_#.test' | tr # $i
done

tr unfortunately does not work for more than one character, such as when you want to substitute # with 10. You are better off using sed for that reason.

for i in 1..10
do
 echo '_#.test' | sed "s/#/$i/"
done

edited Aug 28 at 13:13

answered Aug 27 at 18:26

unxnut

3,3802918

1

What about the "10" result?
â€“Â RudiC
Aug 28 at 8:02

Thanks @RudiC. Fixed my answer.
â€“Â unxnut
Aug 28 at 13:13

add a commentÂ |Â

up vote
1
down vote

I would do it with parameter expansion:

$ var='test#asdf'
$ for i in 1..10; do echo "$var/#/"$i""; done
test1asdf
test2asdf
test3asdf
test4asdf
test5asdf
test6asdf
test7asdf
test8asdf
test9asdf
test10asdf

The $parameter/pattern/string expansion

takes the expansion of $parameter (in our case, $var) and

replaces the first occurrence of pattern (the escaped # â€“ /# has a special meaning in the context, "replace at the beginning of the string", which we want to avoid) with

string ("$i" in our case)

Alternatively, you could replace the # with %d and use it as the format string for printf:

printf "$var/#/%d\n" 1..10

answered Aug 27 at 18:52

Benjamin W.

363110

add a commentÂ |Â

up vote
0
down vote

If what you need is a string expansion, then this is enough:

echo 'test'1..11'asdf'
test1asdf test2asdf test3asdf test4asdf test5asdf test6asdf test7asdf test8asdf test9asdf test10asdf test11asdf

If you need to replace a # in several strings, you may use the positional arguments:

$ set -- test#asdf,,,,,,,,,,
$ printf '%s ' "$@"; echo
test#asdf test#asdf test#asdf test#asdf test#asdf test#asdf test#asdf test#asdf test#asdf

$ j=0; for i do echo "$i//#/"$((j+=1))""; done
test1asdf
test2asdf
test3asdf
test4asdf
test5asdf
test6asdf
test7asdf
test8asdf
test9asdf
test10asdf
test11asdf

answered Aug 27 at 22:58

Isaac

6,8001834

add a commentÂ |Â

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6 Answers
6

active

oldest

votes

6 Answers
6

active

oldest

votes

up vote
6
down vote

accepted

The 0..10 zsh operator (now also supported by a few other shells including bash) is just another form of csh-style brace expansion.

It is expanded by the shell before calling the command. The command doesn't see those 0..10.

Here, you'd want:

echo '_'0..10'.test'

for echo to be passed "_0.test", "_1.test", ..., "_10.test" as arguments.

Or if you wanted that # to be translated into that 0..10 operator, transform it into shell code to be evaluated:

eval "$(echo 'echo _#.test' | sed 's/#/0..10/')"

where eval is being passed echo _0..10.test as arguments.

(not that I would recommend doing anything like that).

edited Aug 27 at 19:24

answered Aug 27 at 18:32

StÃ©phane Chazelas

283k53521855

add a commentÂ |Â

up vote
6
down vote

accepted

The 0..10 zsh operator (now also supported by a few other shells including bash) is just another form of csh-style brace expansion.

It is expanded by the shell before calling the command. The command doesn't see those 0..10.

Here, you'd want:

echo '_'0..10'.test'

for echo to be passed "_0.test", "_1.test", ..., "_10.test" as arguments.

Or if you wanted that # to be translated into that 0..10 operator, transform it into shell code to be evaluated:

eval "$(echo 'echo _#.test' | sed 's/#/0..10/')"

where eval is being passed echo _0..10.test as arguments.

(not that I would recommend doing anything like that).

edited Aug 27 at 19:24

answered Aug 27 at 18:32

StÃ©phane Chazelas

283k53521855

add a commentÂ |Â

up vote
6
down vote

accepted

The 0..10 zsh operator (now also supported by a few other shells including bash) is just another form of csh-style brace expansion.

It is expanded by the shell before calling the command. The command doesn't see those 0..10.

Here, you'd want:

echo '_'0..10'.test'

for echo to be passed "_0.test", "_1.test", ..., "_10.test" as arguments.

Or if you wanted that # to be translated into that 0..10 operator, transform it into shell code to be evaluated:

eval "$(echo 'echo _#.test' | sed 's/#/0..10/')"

where eval is being passed echo _0..10.test as arguments.

(not that I would recommend doing anything like that).

edited Aug 27 at 19:24

answered Aug 27 at 18:32

StÃ©phane Chazelas

283k53521855

The 0..10 zsh operator (now also supported by a few other shells including bash) is just another form of csh-style brace expansion.

It is expanded by the shell before calling the command. The command doesn't see those 0..10.

Here, you'd want:

echo '_'0..10'.test'

for echo to be passed "_0.test", "_1.test", ..., "_10.test" as arguments.

Or if you wanted that # to be translated into that 0..10 operator, transform it into shell code to be evaluated:

eval "$(echo 'echo _#.test' | sed 's/#/0..10/')"

where eval is being passed echo _0..10.test as arguments.

(not that I would recommend doing anything like that).

edited Aug 27 at 19:24

answered Aug 27 at 18:32

StÃ©phane Chazelas

283k53521855

edited Aug 27 at 19:24

answered Aug 27 at 18:32

StÃ©phane Chazelas

283k53521855

answered Aug 27 at 18:32

StÃ©phane Chazelas

283k53521855

answered Aug 27 at 18:32

StÃ©phane Chazelas

283k53521855

add a commentÂ |Â

up vote
6
down vote

You can split the string on the delimiter, capture the prefix and the suffix, then use brace expansion to generate the names:

str='test#asdf'
IFS='#' read -r prefix suffix <<<"$str"
names=( "$prefix"0..10"$suffix" )
declare -p names

declare -a names='([0]="test0asdf" [1]="test1asdf" [2]="test2asdf" [3]="test3asdf" [4]="test4asdf" [5]="test5asdf" [6]="test6asdf" [7]="test7asdf" [8]="test8asdf" [9]="test9asdf" [10]="test10asdf")'

edited Aug 27 at 20:42

answered Aug 27 at 19:07

glenn jackman

47k265103

add a commentÂ |Â

up vote
6
down vote

You can split the string on the delimiter, capture the prefix and the suffix, then use brace expansion to generate the names:

str='test#asdf'
IFS='#' read -r prefix suffix <<<"$str"
names=( "$prefix"0..10"$suffix" )
declare -p names

declare -a names='([0]="test0asdf" [1]="test1asdf" [2]="test2asdf" [3]="test3asdf" [4]="test4asdf" [5]="test5asdf" [6]="test6asdf" [7]="test7asdf" [8]="test8asdf" [9]="test9asdf" [10]="test10asdf")'

edited Aug 27 at 20:42

answered Aug 27 at 19:07

glenn jackman

47k265103

add a commentÂ |Â

up vote
6
down vote

You can split the string on the delimiter, capture the prefix and the suffix, then use brace expansion to generate the names:

str='test#asdf'
IFS='#' read -r prefix suffix <<<"$str"
names=( "$prefix"0..10"$suffix" )
declare -p names

declare -a names='([0]="test0asdf" [1]="test1asdf" [2]="test2asdf" [3]="test3asdf" [4]="test4asdf" [5]="test5asdf" [6]="test6asdf" [7]="test7asdf" [8]="test8asdf" [9]="test9asdf" [10]="test10asdf")'

edited Aug 27 at 20:42

answered Aug 27 at 19:07

glenn jackman

47k265103

You can split the string on the delimiter, capture the prefix and the suffix, then use brace expansion to generate the names:

str='test#asdf'
IFS='#' read -r prefix suffix <<<"$str"
names=( "$prefix"0..10"$suffix" )
declare -p names

declare -a names='([0]="test0asdf" [1]="test1asdf" [2]="test2asdf" [3]="test3asdf" [4]="test4asdf" [5]="test5asdf" [6]="test6asdf" [7]="test7asdf" [8]="test8asdf" [9]="test9asdf" [10]="test10asdf")'

edited Aug 27 at 20:42

answered Aug 27 at 19:07

glenn jackman

47k265103

edited Aug 27 at 20:42

answered Aug 27 at 19:07

glenn jackman

47k265103

answered Aug 27 at 19:07

glenn jackman

47k265103

answered Aug 27 at 19:07

glenn jackman

47k265103

add a commentÂ |Â

up vote
2
down vote

Do you have to use #? Maybe you could use %d?

$ for i in 1..10; do printf "_%d.test " "$i"; done
_1.test _2.test _3.test _4.test _5.test _6.test _7.test _8.test _9.test _10.test

answered Aug 27 at 18:29

DopeGhoti

40.7k54979

add a commentÂ |Â

up vote
2
down vote

Do you have to use #? Maybe you could use %d?

$ for i in 1..10; do printf "_%d.test " "$i"; done
_1.test _2.test _3.test _4.test _5.test _6.test _7.test _8.test _9.test _10.test

answered Aug 27 at 18:29

DopeGhoti

40.7k54979

add a commentÂ |Â

up vote
2
down vote

Do you have to use #? Maybe you could use %d?

$ for i in 1..10; do printf "_%d.test " "$i"; done
_1.test _2.test _3.test _4.test _5.test _6.test _7.test _8.test _9.test _10.test

answered Aug 27 at 18:29

DopeGhoti

40.7k54979

Do you have to use #? Maybe you could use %d?

$ for i in 1..10; do printf "_%d.test " "$i"; done
_1.test _2.test _3.test _4.test _5.test _6.test _7.test _8.test _9.test _10.test

answered Aug 27 at 18:29

DopeGhoti

40.7k54979

answered Aug 27 at 18:29

DopeGhoti

40.7k54979

answered Aug 27 at 18:29

DopeGhoti

40.7k54979

answered Aug 27 at 18:29

DopeGhoti

40.7k54979

add a commentÂ |Â

up vote
2
down vote

First # starts a comment. So, you need to escape it with .

Second, use a for loop.

Here is your solution:

for i in 1..10
do
 echo '_#.test' | tr # $i
done

tr unfortunately does not work for more than one character, such as when you want to substitute # with 10. You are better off using sed for that reason.

for i in 1..10
do
 echo '_#.test' | sed "s/#/$i/"
done

edited Aug 28 at 13:13

answered Aug 27 at 18:26

unxnut

3,3802918

1

What about the "10" result?
â€“Â RudiC
Aug 28 at 8:02

Thanks @RudiC. Fixed my answer.
â€“Â unxnut
Aug 28 at 13:13

add a commentÂ |Â

up vote
2
down vote

First # starts a comment. So, you need to escape it with .

Second, use a for loop.

Here is your solution:

for i in 1..10
do
 echo '_#.test' | tr # $i
done

tr unfortunately does not work for more than one character, such as when you want to substitute # with 10. You are better off using sed for that reason.

for i in 1..10
do
 echo '_#.test' | sed "s/#/$i/"
done

edited Aug 28 at 13:13

answered Aug 27 at 18:26

unxnut

3,3802918

1

What about the "10" result?
â€“Â RudiC
Aug 28 at 8:02

Thanks @RudiC. Fixed my answer.
â€“Â unxnut
Aug 28 at 13:13

add a commentÂ |Â

up vote
2
down vote

First # starts a comment. So, you need to escape it with .

Second, use a for loop.

Here is your solution:

for i in 1..10
do
 echo '_#.test' | tr # $i
done

tr unfortunately does not work for more than one character, such as when you want to substitute # with 10. You are better off using sed for that reason.

for i in 1..10
do
 echo '_#.test' | sed "s/#/$i/"
done

edited Aug 28 at 13:13

answered Aug 27 at 18:26

unxnut

3,3802918

First # starts a comment. So, you need to escape it with .

Second, use a for loop.

Here is your solution:

for i in 1..10
do
 echo '_#.test' | tr # $i
done

tr unfortunately does not work for more than one character, such as when you want to substitute # with 10. You are better off using sed for that reason.

for i in 1..10
do
 echo '_#.test' | sed "s/#/$i/"
done

edited Aug 28 at 13:13

answered Aug 27 at 18:26

unxnut

3,3802918

edited Aug 28 at 13:13

answered Aug 27 at 18:26

unxnut

3,3802918

answered Aug 27 at 18:26

unxnut

3,3802918

answered Aug 27 at 18:26

unxnut

3,3802918

1

What about the "10" result?
â€“Â RudiC
Aug 28 at 8:02

Thanks @RudiC. Fixed my answer.
â€“Â unxnut
Aug 28 at 13:13

add a commentÂ |Â

1

What about the "10" result?
â€“Â RudiC
Aug 28 at 8:02

Thanks @RudiC. Fixed my answer.
â€“Â unxnut
Aug 28 at 13:13

What about the "10" result?
â€“Â RudiC
Aug 28 at 8:02

Thanks @RudiC. Fixed my answer.
â€“Â unxnut
Aug 28 at 13:13

add a commentÂ |Â

up vote
1
down vote

I would do it with parameter expansion:

$ var='test#asdf'
$ for i in 1..10; do echo "$var/#/"$i""; done
test1asdf
test2asdf
test3asdf
test4asdf
test5asdf
test6asdf
test7asdf
test8asdf
test9asdf
test10asdf

The $parameter/pattern/string expansion

takes the expansion of $parameter (in our case, $var) and

replaces the first occurrence of pattern (the escaped # â€“ /# has a special meaning in the context, "replace at the beginning of the string", which we want to avoid) with

string ("$i" in our case)

Alternatively, you could replace the # with %d and use it as the format string for printf:

printf "$var/#/%d\n" 1..10

answered Aug 27 at 18:52

Benjamin W.

363110

add a commentÂ |Â

up vote
1
down vote

I would do it with parameter expansion:

$ var='test#asdf'
$ for i in 1..10; do echo "$var/#/"$i""; done
test1asdf
test2asdf
test3asdf
test4asdf
test5asdf
test6asdf
test7asdf
test8asdf
test9asdf
test10asdf

The $parameter/pattern/string expansion

takes the expansion of $parameter (in our case, $var) and

replaces the first occurrence of pattern (the escaped # â€“ /# has a special meaning in the context, "replace at the beginning of the string", which we want to avoid) with

string ("$i" in our case)

Alternatively, you could replace the # with %d and use it as the format string for printf:

printf "$var/#/%d\n" 1..10

answered Aug 27 at 18:52

Benjamin W.

363110

add a commentÂ |Â

up vote
1
down vote

I would do it with parameter expansion:

$ var='test#asdf'
$ for i in 1..10; do echo "$var/#/"$i""; done
test1asdf
test2asdf
test3asdf
test4asdf
test5asdf
test6asdf
test7asdf
test8asdf
test9asdf
test10asdf

The $parameter/pattern/string expansion

takes the expansion of $parameter (in our case, $var) and

replaces the first occurrence of pattern (the escaped # â€“ /# has a special meaning in the context, "replace at the beginning of the string", which we want to avoid) with

string ("$i" in our case)

Alternatively, you could replace the # with %d and use it as the format string for printf:

printf "$var/#/%d\n" 1..10

answered Aug 27 at 18:52

Benjamin W.

363110

I would do it with parameter expansion:

$ var='test#asdf'
$ for i in 1..10; do echo "$var/#/"$i""; done
test1asdf
test2asdf
test3asdf
test4asdf
test5asdf
test6asdf
test7asdf
test8asdf
test9asdf
test10asdf

The $parameter/pattern/string expansion

takes the expansion of $parameter (in our case, $var) and

replaces the first occurrence of pattern (the escaped # â€“ /# has a special meaning in the context, "replace at the beginning of the string", which we want to avoid) with

string ("$i" in our case)

Alternatively, you could replace the # with %d and use it as the format string for printf:

printf "$var/#/%d\n" 1..10

answered Aug 27 at 18:52

Benjamin W.

363110

answered Aug 27 at 18:52

Benjamin W.

363110

answered Aug 27 at 18:52

Benjamin W.

363110

answered Aug 27 at 18:52

Benjamin W.

363110

add a commentÂ |Â

up vote
0
down vote

If what you need is a string expansion, then this is enough:

echo 'test'1..11'asdf'
test1asdf test2asdf test3asdf test4asdf test5asdf test6asdf test7asdf test8asdf test9asdf test10asdf test11asdf

If you need to replace a # in several strings, you may use the positional arguments:

$ set -- test#asdf,,,,,,,,,,
$ printf '%s ' "$@"; echo
test#asdf test#asdf test#asdf test#asdf test#asdf test#asdf test#asdf test#asdf test#asdf

$ j=0; for i do echo "$i//#/"$((j+=1))""; done
test1asdf
test2asdf
test3asdf
test4asdf
test5asdf
test6asdf
test7asdf
test8asdf
test9asdf
test10asdf
test11asdf

answered Aug 27 at 22:58

Isaac

6,8001834

add a commentÂ |Â

up vote
0
down vote

If what you need is a string expansion, then this is enough:

echo 'test'1..11'asdf'
test1asdf test2asdf test3asdf test4asdf test5asdf test6asdf test7asdf test8asdf test9asdf test10asdf test11asdf

If you need to replace a # in several strings, you may use the positional arguments:

$ set -- test#asdf,,,,,,,,,,
$ printf '%s ' "$@"; echo
test#asdf test#asdf test#asdf test#asdf test#asdf test#asdf test#asdf test#asdf test#asdf

$ j=0; for i do echo "$i//#/"$((j+=1))""; done
test1asdf
test2asdf
test3asdf
test4asdf
test5asdf
test6asdf
test7asdf
test8asdf
test9asdf
test10asdf
test11asdf

answered Aug 27 at 22:58

Isaac

6,8001834

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If what you need is a string expansion, then this is enough:

echo 'test'1..11'asdf'
test1asdf test2asdf test3asdf test4asdf test5asdf test6asdf test7asdf test8asdf test9asdf test10asdf test11asdf

If you need to replace a # in several strings, you may use the positional arguments:

$ set -- test#asdf,,,,,,,,,,
$ printf '%s ' "$@"; echo
test#asdf test#asdf test#asdf test#asdf test#asdf test#asdf test#asdf test#asdf test#asdf

$ j=0; for i do echo "$i//#/"$((j+=1))""; done
test1asdf
test2asdf
test3asdf
test4asdf
test5asdf
test6asdf
test7asdf
test8asdf
test9asdf
test10asdf
test11asdf

answered Aug 27 at 22:58

Isaac

6,8001834

If what you need is a string expansion, then this is enough:

echo 'test'1..11'asdf'
test1asdf test2asdf test3asdf test4asdf test5asdf test6asdf test7asdf test8asdf test9asdf test10asdf test11asdf

If you need to replace a # in several strings, you may use the positional arguments:

$ set -- test#asdf,,,,,,,,,,
$ printf '%s ' "$@"; echo
test#asdf test#asdf test#asdf test#asdf test#asdf test#asdf test#asdf test#asdf test#asdf

$ j=0; for i do echo "$i//#/"$((j+=1))""; done
test1asdf
test2asdf
test3asdf
test4asdf
test5asdf
test6asdf
test7asdf
test8asdf
test9asdf
test10asdf
test11asdf

answered Aug 27 at 22:58

Isaac

6,8001834

answered Aug 27 at 22:58

Isaac

6,8001834

answered Aug 27 at 22:58

Isaac

6,8001834

answered Aug 27 at 22:58

Isaac

6,8001834

add a commentÂ |Â

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