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.
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 '#'.
n = int(input())
matriz = []

i = 0
while i < n:
    linha = input()
    matriz.append(linha)
    i += 1


dic = dict()
resposta = 1

i = 0
j = 0

while i < n:
    while j < n:
        if matriz[i][j] == '#':
            if (i,j) not in dic:
                
                if (j-1)>=0 and (j+1)<n and (i+2)<n:
                    num1 = matriz[i+1][j]
                    num2 = matriz[i+1][j-1]
                    num3 = matriz[i+1][j+1]
                    num4 = matriz[i+2][j]
                    if (num1=='#'):
                        if (i+1,j) not in dic:
                            dic[(i+1,j)] = 1
                        else:
                            resposta = 0
                            break
                    else:
                        resposta = 0
                        break

                    if (num2=='#'):
                        if (i+1,j-1) not in dic:
                            dic[(i+1,j-1)] = 1
                        else:
                            resposta = 0
                            break
                    else:
                        resposta = 0
                        break

                    if (num3=='#'):
                        if (i+1,j+1) not in dic:
                            dic[(i+1,j+1)] = 1
                        else:
                            resposta = 0
                            break
                    else:
                        resposta = 0
                        break

                    if (num4=='#'):
                        if (i+2,j) not in dic:
                            dic[(i+2,j)] = 1
                        else:
                            resposta = 0
                            break
                    else:
                        resposta = 0
                        break

                    
                else:
                    resposta = 0
                    break
            
        j += 1
    
    if resposta == 0:
        break
    i += 1

if resposta == 1:
    print("YES")
else:
    print("NO")
	 		 	 	  	 					    	  	 			