Problem: You are given a grid of squares with H horizontal rows and W vertical columns, where each square is painted white or black. HW characters from A_{11} to A_{HW} represent the colors of the squares. A_{ij} is `#` if the square at the i-th row from the top and the j-th column from the left is black, and A_{ij} is `.` if that square is white.

We will repeatedly perform the following operation until all the squares are black:

* Every white square that shares a side with a black square, becomes black.



Find the number of operations that will be performed. The initial grid has at least one black square.

Constraints

* 1 \leq H,W \leq 1000
* A_{ij} is `#` or `.`.
* The given grid has at least one black square.

Input

Input is given from Standard Input in the following format:


H W
A_{11}A_{12}...A_{1W}
:
A_{H1}A_{H2}...A_{HW}


Output

Print the number of operations that will be performed.

Examples

Input

3 3
...
.#.
...


Output

2


Input

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


Output

3

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;

int dx[] = {-1, 0, 0, 1};
int dy[] = {0, -1, 1, 0};

char s[1005][1005];
int d[1005][1005];

int main() {
   int n, m;
   scanf("%d %d", &n, &m);
   queue<pair<int, int>> q;
   for (int i = 1; i <= n; ++i) {
      scanf("%s", s[i] + 1);
      for (int j = 1; j <= m; ++j) {
         if (s[i][j] == '.') d[i][j] = -1;
         else { d[i][j] = 0; q.push({i, j}); }
      }
   }
   int ans = 0;
   while (!q.empty()) {
      int x = q.front().first, y = q.front().second;
      q.pop();
      ans = max(ans, d[x][y]);
      for (int i = 0; i < 4; ++i) {
         int xx = x + dx[i], yy = y + dy[i];
         if (xx <= 0 || xx > n || yy <= 0 || yy > m) continue;
         if (d[xx][yy] == -1) {
            d[xx][yy] = d[x][y] + 1;
            q.push({xx, yy});
         }
      }
   }
   printf("%d\n", ans);
}