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;
struct pt{
  int x,y,n;
};
vector<pt> dirs = { {0,1,1}, {0,-1,1}, {1,0,1}, {-1,0,1} };
int main(){
  ios::sync_with_stdio(0);
  cin.tie(0);
  int h,w,nb = 0;
  cin >> h >> w;
  vector<string> vs(h);
  queue<pt> q;
  for(int i = 0; i < h; ++i){
    cin >> vs[i];
    for(int j = 0; j < w; ++j){
      if(vs[i][j] == '#'){
        vs[i][j] = true;
        q.emplace(pt{i,j,0});
      }
    }
  }
  while(!q.empty()){
    auto p = q.front();
    q.pop();
    for(auto& d : dirs){
      int x=p.x+d.x,y=p.y+d.y,n=p.n+d.n;
      if(0 <= x and x < h and 0 <= y and y < w){
        if(vs[x][y] == '.'){
          vs[x][y] = '#';
          nb = max(nb,n);
          q.emplace(pt{x,y,n});
        }
      }
    }
  }
  cout << nb;
  return 0;
}