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>
using namespace std;
typedef long long ll;
typedef pair <int,int> P;
    
int main() {
  int h,w;
  char b[1010][1010];
  int cnt[1010][1010];
  cin>>h>>w;
  queue<P> que;
  for (int i=0; i<h; i++) {
    for (int j=0; j<w; j++) {
      cin>>b[i][j];
      cnt[i][j]=-1;
      if (b[i][j]=='#') {
	cnt[i][j]=0;
	que.push(P(i,j));
      }
    }
  }

  int dx[4]={1,0,-1,0};
  int dy[4]={0,1,0,-1};
  int ans;
  while (que.size()) {
    P p=que.front();
    que.pop();
    int x=p.first,y=p.second;
    for (int i=0; i<4; i++) {
      if (x+dx[i]>=0&&x+dx[i]<h&&y+dy[i]>=0&&y+dy[i]<w&&b[x+dx[i]][y+dy[i]]!='#') {
	b[x+dx[i]][y+dy[i]]='#';
	cnt[x+dx[i]][y+dy[i]]=cnt[x][y]+1;
	que.push(P(x+dx[i],y+dy[i]));
	ans=cnt[x][y]+1;
      }
    }
  }

  cout<<ans<<endl;
  return 0;
}