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>
#define P pair<int, int>

using namespace std;
const int INF=1e5;

int dx[4]={-1,0,1,0};
int dy[4]={0,-1,0,1};


int main(){
  int h,w,a[1009][1009];
  string s[1009];
  cin >> h >> w;
  memset(a, -1, sizeof(a));
  for(int i=0;i<h;i++)cin >> s[i];
  queue<P> q;
  for(int i=0;i<h;i++){
    for(int j=0;j<w;j++){
      if(s[i][j]=='#'){
        a[i][j]=0;
        q.push(P(i,j));
      }
    }
  }
  int mx=0;
  while(!q.empty()){
    P p=q.front();q.pop();
    int ci=p.first;
    int cj=p.second;
    for(int i=0;i<4;i++){
      int ni=ci+dy[i];
      int nj=cj+dx[i];
      if(!(0<=ni&&ni<h&&0<=nj&&nj<w))continue;
      if(a[ni][nj]==-1){
        a[ni][nj]=a[ci][cj]+1;
        mx = max(mx, a[ni][nj]);
        q.push(P(ni, nj));
      }
    }
  }
  cout << mx << endl;
}