List value swapping: What's the correct order and why?
For a list of integers, such as A = [2, 10, -5], I get the error
Traceback (most recent call last):
File "so.py", line 6, in <module>
v, A[v-1] = A[v-1], v
IndexError: list assignment index out of range
Code:
for i, v in enumerate(A):
while 1<=v<=len(A) and v != A[v-1]:
v, A[v-1] = A[v-1], v
but this works:
for i, v in enumerate(A):
while 1<=v<=len(A) and v != A[v-1]:
A[v-1], v = v, A[v-1]
Why the order of the swapping elements matters here? v is always being checked to be in bound.
Weirdly enough cannot reproduce a smaller example. But,
A = [6, 5, 4, 3, 2]
becomes an infinite loop.
python
edited Nov 15 '18 at 18:19
martineau
69.1k1092186
asked Nov 15 '18 at 17:22
kaiserasifkaiserasif
588
|
show 3 more comments
For a list of integers, such as A = [2, 10, -5], I get the error
Traceback (most recent call last):
File "so.py", line 6, in <module>
v, A[v-1] = A[v-1], v
IndexError: list assignment index out of range
Code:
for i, v in enumerate(A):
while 1<=v<=len(A) and v != A[v-1]:
v, A[v-1] = A[v-1], v
but this works:
for i, v in enumerate(A):
while 1<=v<=len(A) and v != A[v-1]:
A[v-1], v = v, A[v-1]
Why the order of the swapping elements matters here? v is always being checked to be in bound.
Weirdly enough cannot reproduce a smaller example. But,
A = [6, 5, 4, 3, 2]
becomes an infinite loop.
python
edited Nov 15 '18 at 18:19
martineau
69.1k1092186
asked Nov 15 '18 at 17:22
kaiserasifkaiserasif
588
4
Can you provide A and n?
– Ssein
Nov 15 '18 at 17:29
1
You should refer to this post.
– Dimeji Mudele
Nov 15 '18 at 17:41
Don't reassign the iteration variable,v.
– hpaulj
Nov 15 '18 at 17:55
1
iis the index.vthe value.A[i]makes sense.A[v]does not.
– hpaulj
Nov 15 '18 at 17:58
Note that your inner loop is awhile, not anif. Your list [6, 5, 4, 3, 2] gets stuck in thewhile.
– Prune
Nov 15 '18 at 18:10
|
show 3 more comments
For a list of integers, such as A = [2, 10, -5], I get the error
Traceback (most recent call last):
File "so.py", line 6, in <module>
v, A[v-1] = A[v-1], v
IndexError: list assignment index out of range
Code:
for i, v in enumerate(A):
while 1<=v<=len(A) and v != A[v-1]:
v, A[v-1] = A[v-1], v
but this works:
for i, v in enumerate(A):
while 1<=v<=len(A) and v != A[v-1]:
A[v-1], v = v, A[v-1]
Why the order of the swapping elements matters here? v is always being checked to be in bound.
Weirdly enough cannot reproduce a smaller example. But,
A = [6, 5, 4, 3, 2]
becomes an infinite loop.
python
edited Nov 15 '18 at 18:19
martineau
69.1k1092186
asked Nov 15 '18 at 17:22
kaiserasifkaiserasif
588
For a list of integers, such as A = [2, 10, -5], I get the error
Traceback (most recent call last):
File "so.py", line 6, in <module>
v, A[v-1] = A[v-1], v
IndexError: list assignment index out of range
Code:
for i, v in enumerate(A):
while 1<=v<=len(A) and v != A[v-1]:
v, A[v-1] = A[v-1], v
but this works:
for i, v in enumerate(A):
while 1<=v<=len(A) and v != A[v-1]:
A[v-1], v = v, A[v-1]
Why the order of the swapping elements matters here? v is always being checked to be in bound.
Weirdly enough cannot reproduce a smaller example. But,
A = [6, 5, 4, 3, 2]
becomes an infinite loop.
python
python
edited Nov 15 '18 at 18:19
martineau
69.1k1092186
asked Nov 15 '18 at 17:22
kaiserasifkaiserasif
588
edited Nov 15 '18 at 18:19
martineau
69.1k1092186
asked Nov 15 '18 at 17:22
kaiserasifkaiserasif
588
edited Nov 15 '18 at 18:19
martineau
69.1k1092186
edited Nov 15 '18 at 18:19
martineau
69.1k1092186
edited Nov 15 '18 at 18:19
martineau
69.1k1092186
69.1k1092186
asked Nov 15 '18 at 17:22
kaiserasifkaiserasif
588
asked Nov 15 '18 at 17:22
kaiserasifkaiserasif
588
asked Nov 15 '18 at 17:22
kaiserasifkaiserasif
588
588
4
Can you provide A and n?
– Ssein
Nov 15 '18 at 17:29
1
You should refer to this post.
– Dimeji Mudele
Nov 15 '18 at 17:41
Don't reassign the iteration variable,v.
– hpaulj
Nov 15 '18 at 17:55
1
iis the index.vthe value.A[i]makes sense.A[v]does not.
– hpaulj
Nov 15 '18 at 17:58
Note that your inner loop is awhile, not anif. Your list [6, 5, 4, 3, 2] gets stuck in thewhile.
– Prune
Nov 15 '18 at 18:10
|
show 3 more comments
4
Can you provide A and n?
– Ssein
Nov 15 '18 at 17:29
1
You should refer to this post.
– Dimeji Mudele
Nov 15 '18 at 17:41
Don't reassign the iteration variable,v.
– hpaulj
Nov 15 '18 at 17:55
1
iis the index.vthe value.A[i]makes sense.A[v]does not.
– hpaulj
Nov 15 '18 at 17:58
Note that your inner loop is awhile, not anif. Your list [6, 5, 4, 3, 2] gets stuck in thewhile.
– Prune
Nov 15 '18 at 18:10
4
4
Can you provide A and n?
– Ssein
Nov 15 '18 at 17:29
Can you provide A and n?
– Ssein
Nov 15 '18 at 17:29
1
1
You should refer to this post.
– Dimeji Mudele
Nov 15 '18 at 17:41
You should refer to this post.
– Dimeji Mudele
Nov 15 '18 at 17:41
Don't reassign the iteration variable,
v.– hpaulj
Nov 15 '18 at 17:55
Don't reassign the iteration variable,
v.– hpaulj
Nov 15 '18 at 17:55
1
1
i is the index. v the value. A[i] makes sense. A[v] does not.– hpaulj
Nov 15 '18 at 17:58
i is the index. v the value. A[i] makes sense. A[v] does not.– hpaulj
Nov 15 '18 at 17:58
Note that your inner loop is a
while, not an if. Your list [6, 5, 4, 3, 2] gets stuck in the while.– Prune
Nov 15 '18 at 18:10
Note that your inner loop is a
while, not an if. Your list [6, 5, 4, 3, 2] gets stuck in the while.– Prune
Nov 15 '18 at 18:10
|
show 3 more comments
3 Answers
3
active
oldest
votes
I reproduced this with [2, 10, -5].
The detailed sequence of operations is
i, v = 0, 2
1 <= 2 ? OK
2 != A[1] ? OK ... stay in loop
v, A[v-1] = 10, 2
# This assignment breaks into ...
v = 10
A[10-1] = 2
# ... and this second assignment is out of range.
If you switch the assignment order:
A[v-1], v = 2, 10
# This assignment breaks into ...
A[2-1] = 2
v = 10
In this case, your while conditions have properly guarded a legal assignment.
Note that you are not swapping list values; v is a local variable; it is not a reference to A[i]. For instance, in the above example, you do get v=10, but this does not affect the value of A[0].
answered Nov 15 '18 at 17:56
PrunePrune
45.2k143559
add a comment |
Python swaps the variables in the order provided, so v is assigned the value at A[v-1], and then tries to reassign A[v-1] - but since v has been modified to be a list element, v-1 is out of range of A.
answered Nov 15 '18 at 17:35
TimTim
1,772621
Then, why the opposite doesn't cause the problem, i.e.A[v-1]set tovfirst, thenvstill gets the previousA[v-1]value from before the update.
– kaiserasif
Nov 15 '18 at 17:44
1
Because you have an if statement ensuring thatvis less than the length of the array, soA[v-1]is always set to a value that can indexAin the next step.
– Tim
Nov 15 '18 at 17:57
add a comment |
A = [6, 5, 4, 3, 2]
for i, v in enumerate(A):
while 1<=v<=len(A) and v != A[v-1]:
v, A[v-1] = A[v-1], v
You need to try running out your algorithm in your head:
Start:
i = 0, v = 6
v(6) is not between 1 and 5: next iteration
i = 1, v = 5, A[5-1] = 2
5 is between 1 and 5; v is not equal to 2: swap ->
v = 2; A[4] = 2
i = 1, v = 2, A[2-1] = 5
2 is between 1 and 5; v is not equal to 5: swap ->
v = 5; A[1] = 5
i = 1, v = 5, A[5-1] = 2
5 is between 1 and 5; v is not equal to 2: swap ->
v = 2; A[4] = 2
i = 1, v = 2, A[2-1] = 5
2 is between 1 and 5; v is not equal to 5: swap ->
v = 5; A[1] = 5
... and on and on
I don't think your algorithm makes sense. It's unclear why you are using the values in your list to index the list during your loop. I think this confusion about the index and values is at the root of your problem.
answered Nov 15 '18 at 18:07
vaer-kvaer-k
3,94852335
add a comment |
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3 Answers
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oldest
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3 Answers
3
active
oldest
votes
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oldest
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active
oldest
votes
I reproduced this with [2, 10, -5].
The detailed sequence of operations is
i, v = 0, 2
1 <= 2 ? OK
2 != A[1] ? OK ... stay in loop
v, A[v-1] = 10, 2
# This assignment breaks into ...
v = 10
A[10-1] = 2
# ... and this second assignment is out of range.
If you switch the assignment order:
A[v-1], v = 2, 10
# This assignment breaks into ...
A[2-1] = 2
v = 10
In this case, your while conditions have properly guarded a legal assignment.
Note that you are not swapping list values; v is a local variable; it is not a reference to A[i]. For instance, in the above example, you do get v=10, but this does not affect the value of A[0].
answered Nov 15 '18 at 17:56
PrunePrune
45.2k143559
add a comment |
I reproduced this with [2, 10, -5].
The detailed sequence of operations is
i, v = 0, 2
1 <= 2 ? OK
2 != A[1] ? OK ... stay in loop
v, A[v-1] = 10, 2
# This assignment breaks into ...
v = 10
A[10-1] = 2
# ... and this second assignment is out of range.
If you switch the assignment order:
A[v-1], v = 2, 10
# This assignment breaks into ...
A[2-1] = 2
v = 10
In this case, your while conditions have properly guarded a legal assignment.
Note that you are not swapping list values; v is a local variable; it is not a reference to A[i]. For instance, in the above example, you do get v=10, but this does not affect the value of A[0].
answered Nov 15 '18 at 17:56
PrunePrune
45.2k143559
add a comment |
I reproduced this with [2, 10, -5].
The detailed sequence of operations is
i, v = 0, 2
1 <= 2 ? OK
2 != A[1] ? OK ... stay in loop
v, A[v-1] = 10, 2
# This assignment breaks into ...
v = 10
A[10-1] = 2
# ... and this second assignment is out of range.
If you switch the assignment order:
A[v-1], v = 2, 10
# This assignment breaks into ...
A[2-1] = 2
v = 10
In this case, your while conditions have properly guarded a legal assignment.
Note that you are not swapping list values; v is a local variable; it is not a reference to A[i]. For instance, in the above example, you do get v=10, but this does not affect the value of A[0].
answered Nov 15 '18 at 17:56
PrunePrune
45.2k143559
I reproduced this with [2, 10, -5].
The detailed sequence of operations is
i, v = 0, 2
1 <= 2 ? OK
2 != A[1] ? OK ... stay in loop
v, A[v-1] = 10, 2
# This assignment breaks into ...
v = 10
A[10-1] = 2
# ... and this second assignment is out of range.
If you switch the assignment order:
A[v-1], v = 2, 10
# This assignment breaks into ...
A[2-1] = 2
v = 10
In this case, your while conditions have properly guarded a legal assignment.
Note that you are not swapping list values; v is a local variable; it is not a reference to A[i]. For instance, in the above example, you do get v=10, but this does not affect the value of A[0].
answered Nov 15 '18 at 17:56
PrunePrune
45.2k143559
edited Nov 15 '18 at 18:06
answered Nov 15 '18 at 17:56
PrunePrune
45.2k143559
answered Nov 15 '18 at 17:56
PrunePrune
45.2k143559
answered Nov 15 '18 at 17:56
PrunePrune
45.2k143559
45.2k143559
add a comment |
add a comment |
Python swaps the variables in the order provided, so v is assigned the value at A[v-1], and then tries to reassign A[v-1] - but since v has been modified to be a list element, v-1 is out of range of A.
answered Nov 15 '18 at 17:35
TimTim
1,772621
Then, why the opposite doesn't cause the problem, i.e.A[v-1]set tovfirst, thenvstill gets the previousA[v-1]value from before the update.
– kaiserasif
Nov 15 '18 at 17:44
1
Because you have an if statement ensuring thatvis less than the length of the array, soA[v-1]is always set to a value that can indexAin the next step.
– Tim
Nov 15 '18 at 17:57
add a comment |
Python swaps the variables in the order provided, so v is assigned the value at A[v-1], and then tries to reassign A[v-1] - but since v has been modified to be a list element, v-1 is out of range of A.
answered Nov 15 '18 at 17:35
TimTim
1,772621
Then, why the opposite doesn't cause the problem, i.e.A[v-1]set tovfirst, thenvstill gets the previousA[v-1]value from before the update.
– kaiserasif
Nov 15 '18 at 17:44
1
Because you have an if statement ensuring thatvis less than the length of the array, soA[v-1]is always set to a value that can indexAin the next step.
– Tim
Nov 15 '18 at 17:57
add a comment |
Python swaps the variables in the order provided, so v is assigned the value at A[v-1], and then tries to reassign A[v-1] - but since v has been modified to be a list element, v-1 is out of range of A.
answered Nov 15 '18 at 17:35
TimTim
1,772621
Python swaps the variables in the order provided, so v is assigned the value at A[v-1], and then tries to reassign A[v-1] - but since v has been modified to be a list element, v-1 is out of range of A.
answered Nov 15 '18 at 17:35
TimTim
1,772621
answered Nov 15 '18 at 17:35
TimTim
1,772621
answered Nov 15 '18 at 17:35
TimTim
1,772621
answered Nov 15 '18 at 17:35
TimTim
1,772621
1,772621
Then, why the opposite doesn't cause the problem, i.e.A[v-1]set tovfirst, thenvstill gets the previousA[v-1]value from before the update.
– kaiserasif
Nov 15 '18 at 17:44
1
Because you have an if statement ensuring thatvis less than the length of the array, soA[v-1]is always set to a value that can indexAin the next step.
– Tim
Nov 15 '18 at 17:57
add a comment |
Then, why the opposite doesn't cause the problem, i.e.A[v-1]set tovfirst, thenvstill gets the previousA[v-1]value from before the update.
– kaiserasif
Nov 15 '18 at 17:44
1
Because you have an if statement ensuring thatvis less than the length of the array, soA[v-1]is always set to a value that can indexAin the next step.
– Tim
Nov 15 '18 at 17:57
Then, why the opposite doesn't cause the problem, i.e.
A[v-1] set to v first, then v still gets the previous A[v-1] value from before the update.– kaiserasif
Nov 15 '18 at 17:44
Then, why the opposite doesn't cause the problem, i.e.
A[v-1] set to v first, then v still gets the previous A[v-1] value from before the update.– kaiserasif
Nov 15 '18 at 17:44
1
1
Because you have an if statement ensuring that
v is less than the length of the array, so A[v-1] is always set to a value that can index A in the next step.– Tim
Nov 15 '18 at 17:57
Because you have an if statement ensuring that
v is less than the length of the array, so A[v-1] is always set to a value that can index A in the next step.– Tim
Nov 15 '18 at 17:57
add a comment |
A = [6, 5, 4, 3, 2]
for i, v in enumerate(A):
while 1<=v<=len(A) and v != A[v-1]:
v, A[v-1] = A[v-1], v
You need to try running out your algorithm in your head:
Start:
i = 0, v = 6
v(6) is not between 1 and 5: next iteration
i = 1, v = 5, A[5-1] = 2
5 is between 1 and 5; v is not equal to 2: swap ->
v = 2; A[4] = 2
i = 1, v = 2, A[2-1] = 5
2 is between 1 and 5; v is not equal to 5: swap ->
v = 5; A[1] = 5
i = 1, v = 5, A[5-1] = 2
5 is between 1 and 5; v is not equal to 2: swap ->
v = 2; A[4] = 2
i = 1, v = 2, A[2-1] = 5
2 is between 1 and 5; v is not equal to 5: swap ->
v = 5; A[1] = 5
... and on and on
I don't think your algorithm makes sense. It's unclear why you are using the values in your list to index the list during your loop. I think this confusion about the index and values is at the root of your problem.
answered Nov 15 '18 at 18:07
vaer-kvaer-k
3,94852335
add a comment |
A = [6, 5, 4, 3, 2]
for i, v in enumerate(A):
while 1<=v<=len(A) and v != A[v-1]:
v, A[v-1] = A[v-1], v
You need to try running out your algorithm in your head:
Start:
i = 0, v = 6
v(6) is not between 1 and 5: next iteration
i = 1, v = 5, A[5-1] = 2
5 is between 1 and 5; v is not equal to 2: swap ->
v = 2; A[4] = 2
i = 1, v = 2, A[2-1] = 5
2 is between 1 and 5; v is not equal to 5: swap ->
v = 5; A[1] = 5
i = 1, v = 5, A[5-1] = 2
5 is between 1 and 5; v is not equal to 2: swap ->
v = 2; A[4] = 2
i = 1, v = 2, A[2-1] = 5
2 is between 1 and 5; v is not equal to 5: swap ->
v = 5; A[1] = 5
... and on and on
I don't think your algorithm makes sense. It's unclear why you are using the values in your list to index the list during your loop. I think this confusion about the index and values is at the root of your problem.
answered Nov 15 '18 at 18:07
vaer-kvaer-k
3,94852335
add a comment |
A = [6, 5, 4, 3, 2]
for i, v in enumerate(A):
while 1<=v<=len(A) and v != A[v-1]:
v, A[v-1] = A[v-1], v
You need to try running out your algorithm in your head:
Start:
i = 0, v = 6
v(6) is not between 1 and 5: next iteration
i = 1, v = 5, A[5-1] = 2
5 is between 1 and 5; v is not equal to 2: swap ->
v = 2; A[4] = 2
i = 1, v = 2, A[2-1] = 5
2 is between 1 and 5; v is not equal to 5: swap ->
v = 5; A[1] = 5
i = 1, v = 5, A[5-1] = 2
5 is between 1 and 5; v is not equal to 2: swap ->
v = 2; A[4] = 2
i = 1, v = 2, A[2-1] = 5
2 is between 1 and 5; v is not equal to 5: swap ->
v = 5; A[1] = 5
... and on and on
I don't think your algorithm makes sense. It's unclear why you are using the values in your list to index the list during your loop. I think this confusion about the index and values is at the root of your problem.
answered Nov 15 '18 at 18:07
vaer-kvaer-k
3,94852335
A = [6, 5, 4, 3, 2]
for i, v in enumerate(A):
while 1<=v<=len(A) and v != A[v-1]:
v, A[v-1] = A[v-1], v
You need to try running out your algorithm in your head:
Start:
i = 0, v = 6
v(6) is not between 1 and 5: next iteration
i = 1, v = 5, A[5-1] = 2
5 is between 1 and 5; v is not equal to 2: swap ->
v = 2; A[4] = 2
i = 1, v = 2, A[2-1] = 5
2 is between 1 and 5; v is not equal to 5: swap ->
v = 5; A[1] = 5
i = 1, v = 5, A[5-1] = 2
5 is between 1 and 5; v is not equal to 2: swap ->
v = 2; A[4] = 2
i = 1, v = 2, A[2-1] = 5
2 is between 1 and 5; v is not equal to 5: swap ->
v = 5; A[1] = 5
... and on and on
I don't think your algorithm makes sense. It's unclear why you are using the values in your list to index the list during your loop. I think this confusion about the index and values is at the root of your problem.
answered Nov 15 '18 at 18:07
vaer-kvaer-k
3,94852335
answered Nov 15 '18 at 18:07
vaer-kvaer-k
3,94852335
answered Nov 15 '18 at 18:07
vaer-kvaer-k
3,94852335
answered Nov 15 '18 at 18:07
vaer-kvaer-k
3,94852335
3,94852335
add a comment |
add a comment |
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4
Can you provide A and n?
– Ssein
Nov 15 '18 at 17:29
1
You should refer to this post.
– Dimeji Mudele
Nov 15 '18 at 17:41
Don't reassign the iteration variable,
v.– hpaulj
Nov 15 '18 at 17:55
1
iis the index.vthe value.A[i]makes sense.A[v]does not.– hpaulj
Nov 15 '18 at 17:58
Note that your inner loop is a
while, not anif. Your list [6, 5, 4, 3, 2] gets stuck in thewhile.– Prune
Nov 15 '18 at 18:10