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.
#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
std::pair<T, U> operator+(const std::pair<T, U>& l, const std::pair<T, U>& r) {
  return {l.first + r.first, l.second + r.second};
}
typedef void (*callback_function)(void);
const long long INF64 = 1e18;
const long long INF32 = 1e9;
bool visited[102][102];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  {
    long long n;
    cin >> n;
    char a[n + 2][n + 2];
    for (long long i = (0); i <= (n + 1); i++)
      a[i][0] = a[n + 1][0] = a[0][i] = a[0][n + 1] = '.';
    for (long long i = (1); i <= (n); i++)
      for (long long j = (1); j <= (n); j++) cin >> a[i][j];
    long long d[4][2] = {{1, -1}, {1, 0}, {1, 1}, {2, 0}};
    long long x, y;
    string ans = "YES";
    for (long long i = (1); i <= (n); i++) {
      for (long long j = (1); j <= (n); j++) {
        if (a[i][j] == '#' && !visited[i][j]) {
          visited[i][j] = true;
          for (long long k = (0); k < (4); k++) {
            x = i + d[k][0], y = j + d[k][1];
            if (a[x][y] != '#' || visited[x][y]) {
              ans = "NO";
              break;
            }
            visited[x][y] = true;
          }
        }
      }
      if (ans != "YES") break;
    }
    cout << ans << endl;
  }
  return 0;
}
