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 '#'.
def cross(m, a, b):
    if len(m) <= a + 1 or a == 0:
        return False
    if len(m[a]) <= b + 1 or b == 0:
        return False
    
    return m[a][b] == '#' and m[a - 1][b] == '#' and m[a + 1][b] == '#' and m[a][b - 1] == '#' and m[a][b + 1] == '#'

def emptyCross(m, a, b):
    m[a][b] = '.'
    m[a - 1][b] = '.'
    m[a + 1][b] = '.'
    m[a][b - 1] = '.'
    m[a][b + 1] = '.'

n = int(input())

m = []

for i in range(n):
    m.append([char for char in input()])

worked = True
for i in range(1, n - 1):
    for j in range(1, n - 1):
        if m[i][j] == '#':
            if cross(m, i, j):
                emptyCross(m, i, j)
    
    if i >= 3:
        if '#' in m[i - 2]:
            worked = False
            break

for i in range(n - 1, n):
    if '#'in m[i]:
        worked = False
        break

if worked:
    print("YES")
else:
    print("NO")
  	  		  	 	  		 	  				 		 		