#228(div2)B. Fox and Cross

B. Fox and Cross

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Fox Ciel has a board with 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.

Ciel wants to draw several (may be zero) crosses on the board. Each cross must cover exactly five cells with symbols '#', 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

The first line contains an integer n (3 ≤ n ≤ 100) — the size of the board.

Each of the next 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

Output a single line with "YES" if Ciel can draw the crosses in the described way. Otherwise output a single line with "NO".

Examples

input

5
.#...
####.
.####
...#.
.....

output

YES

input

4
####
####
####
####

output

NO

input

6
.#....
####..
.####.
.#.##.
######
.#..#.

output

YES

input

6
.#..#.
######
.####.
.####.
######
.#..#.

output

NO

input

3
...
...
...

output

YES
题意：问#是否能组成十字架，不可重复利用
思路：模拟

#include<bits/stdc++.h>
using namespace std;

char s[103][103];

int main(){
    int n;
    cin>>n;
    int sum=0;
    for(int i=1;i<=n;i++){
        scanf("%s",s[i]+1);
        for(int j=1;j<=n;j++){
            if(s[i][j]=='#') sum++;
        }
    }
    if(sum%5!=0) cout<<"NO"<<endl;
    else {
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                if(s[i][j]=='#'){
                    if(s[i+1][j]=='#'&&s[i+2][j]=='#'&&s[i+1][j-1]=='#'&&s[i+1][j+1]=='#'){
                            s[i][j]='.';
                        s[i+1][j]='.';s[i+2][j]='.';s[i+1][j-1]='.';s[i+1][j+1]='.';
                    }
                    else {
                        cout<<"NO"<<endl;return 0;
                    }
                }
            }
        }
        cout<<"YES"<<endl;
    }
}