CF228B题Fox and Cross

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;
}