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 H, W;
string s[1024];

int di[4] = {1, 0, -1, 0};
int dj[4] = {0, -1, 0, 1};

int main(){
  cin >> H >> W;
  for(int i=1;i<=H;i++){
    cin >> s[i];
    s[i] = "*" + s[i] + "*";
  }
  for(int j=0;j<W+2;j++){
    s[0] += '*';
    s[H+1] += '*';
  }
  
  
  queue<int> q;
  for(int i=0;i<=H;i++)for(int j=0;j<=W;j++){
    if (s[i][j]=='#'){
      q.push(i); q.push(j); q.push(0);
    }
  }
  int ans = 0;
  while (!q.empty()){
    int ci = q.front(); q.pop();
    int cj = q.front(); q.pop();
    int ct = q.front(); q.pop();
    ans = max(ans, ct);
    
    for(int i=0;i<4;i++){
      int ni = ci + di[i];
      int nj = cj + dj[i];
      if (s[ni][nj]=='.'){
        s[ni][nj] = '#';
        q.push(ni); q.push(nj); q.push(ct+1);
      }
    }
  }
  cout << ans << endl;
  
  return 0;
}