n rows and n columns. So, the board consists of n × n cells. Each cell contains either a symbol '.', or a symbol '#'.
A cross on the board is a connected set of exactly five cells of the board that looks like a cross. The picture below shows how it looks.
#', and any cell with symbol '#' must belong to some cross. No two crosses can share a cell.
Please, tell Ciel if she can draw the crosses in the described way.
Input
n (3 ≤ n ≤ 100) — the size of the board.
n lines describes one row of the board. The i-th line describes the i-th row of the board and consists of n characters. Each character is either a symbol '.', or a symbol '#'.
Output
YES" if Ciel can draw the crosses in the described way. Otherwise output a single line with "NO".
Sample test(s)
input
5 .#... ####. .#### ...#. .....
output
YES
input
4 #### #### #### ####
output
NO
input
6 .#.... ####.. .####. .#.##. ###### .#..#.
output
YES
input
6 .#..#. ###### .####. .####. ###### .#..#.
output
NO
input
3 ... ... ...
output
YES
Note
In example 1, you can draw two crosses. The picture below shows what they look like.
#', but each cross contains 5. Since 16 is not a multiple of 5, so it's impossible to cover all.
#include <stdio.h>
#include <string.h>
char s[101][101];
int main()
{
int n, i, j, x=0;
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf(" %s",s[i]);
}
for(i=1;i<n-1;i++)
{
for(j=1;j<n-1;j++)
{
if(s[i][j]=='#'&&s[i-1][j]=='#'&&s[i+1][j]=='#'&&s[i][j-1]=='#'&&s[i][j+1]=='#')
{
s[i][j]='.';
s[i-1][j]='.';
s[i+1][j]='.';
s[i][j-1]='.';
s[i][j+1]='.';
}
}
}
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
if(s[i][j]=='#')
{
x=1;
break;
}
}
if(x)
break;
}
if(x)
printf("NO\n");
else
printf("YES\n");
return 0;
}
