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.

<image>

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

Note

In example 1, you can draw two crosses. The picture below shows what they look like.

<image>

In example 2, the board contains 16 cells with '#', but each cross contains 5. Since 16 is not a multiple of 5, so it's impossible to cover all.
import java.util.Scanner;

public class A8 {
	int a[][]=new int[101][101];
	int n=0;
	static int k=0;
	void create(){
		Scanner sc=new Scanner(System.in);
		n=sc.nextInt();
		for(int i=0;i<n;i++){
			String str=sc.next();
			char b[]=str.toCharArray();
			for(int j=0;j<n;j++){
				if(b[j]=='.'){
					a[i][j]=0;
				}else{
					a[i][j]=1;
				}
			}
		}
	}
	void isten(){
		for(int i=1;i<n-1;i++){
			for(int j=1;j<n-1;j++){
				if(a[i][j]==1){
					if(a[i-1][j]==1&&a[i+1][j]==1&&a[i][j-1]==1&&a[i][j+1]==1){
						a[i][j]=0;a[i-1][j]=0;a[i+1][j]=0;a[i][j-1]=0;a[i][j+1]=0;
					}
				}else{
					continue;
				}
			}
		}
	}
	boolean  judge(){
		for(int i=0;i<n;i++){
			for(int j=0;j<n;j++){
				if(a[i][j]==1){
					return false;
				}
			}
		}
		return true;
	}


	public static void main(String[] args) {
		A8 s=new A8();
		s.create();
		s.isten();
		if(s.judge()){
			System.out.println("YES");
		}else{
			System.out.println("NO");
		}

	}

}

		 	 		 			 	  	 	 	   		 		 	