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
#include <bits/stdc++.h>
#include <stdlib.h>
using namespace std;
typedef long long ll;

int main() {
  int h,w;
  cin>>h>>w;
  char a[h][w];
  int res[h][w];
  queue<int> memo;
  for(int i=0;i<h;i++){
    for(int j=0;j<w;j++) {
      cin>>a[i][j];
      if(a[i][j]=='#') {
        res[i][j]=0;
        memo.push(i);memo.push(j);
      }
      else res[i][j]=-1;
    }
  }
  int ans=0;
  while(!memo.empty()){
    int y=memo.front();memo.pop();
    int x=memo.front();memo.pop();
    int dy[4]={0,1,0,-1},dx[4]={1,0,-1,0};
    for(int i=0;i<4;i++){
      int ny=y+dy[i],nx=x+dx[i];
      if(ny<0||ny>=h||nx<0||nx>=w) continue;
      if(res[ny][nx]!=-1) continue;
      res[ny][nx]=res[y][x]+1;
      ans=max(ans,res[ny][nx]);
      memo.push(ny);memo.push(nx);
    }
  }
  cout<<ans<<endl;
  return 0;
}