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,i,j,c = 0,ny,nx,dy,dx,vy[4] = {1,0,-1,0},vx[4] = {0,1,0,-1},s;cin >> h >> w;
  char m[h][w];queue<pair<int,int>> q;
  for(i=0;i<h;i++){
    for(j=0;j<w;j++){
      cin >> m[i][j];
      if(m[i][j]=='#') q.push(make_pair(i,j));
    }
  }
  while(!q.empty()){
    s = q.size();
    for(i=0;i<s;i++){
      ny = q.front().first;nx = q.front().second;
      for(j=0;j<4;j++){
        dy = ny+vy[j];dx = nx+vx[j];
        if(dy>=0 && dx>=0 && dy<h && dx<w && m[dy][dx] == '.'){
          m[dy][dx] = '#';q.push(make_pair(dy,dx));
        }
      }
      q.pop();
    }
    c++;
  }
  cout << c-1 << endl;
}