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 main() {
  int h, w; cin >> h >> w;
  vector<vector<int>> d(h+2, vector<int>(w+2));
  queue<int> q;
  for (int i = 0; i < h; i++) {
    string s; cin >> s;
    for (int j = 0; j < w; j++) {
      if (s[j] == '#') q.push((w+2)*(i+1)+j+1);
      else d[i+1][j+1] = 1e9;
    }
  }
  int res = 0;
  while (q.size()) {
    int p = q.front(); q.pop();
    int r = p / (w+2);
    int c = p % (w+2);
    int dr[] = {-1, 0, 1, 0};
    int dc[] = {0, 1, 0, -1};
    int i = 4;
    while (i--) {
      int nr = r + dr[i];
      int nc = c + dc[i];
      if (d[nr][nc] == 1e9) {
        res = d[nr][nc] = d[r][c] + 1;
        q.push((w+2)*nr+nc);
      }
    }
  }
  cout << res << endl;
}