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;
int dy[4]={0,1,0,-1},dx[4]={1,0,-1,0};
int H,W,ans=0;
char t;
bool maze[1002][1002];
int dist[1002][1002];
queue<pair<int,int>> que;
int main()
{
  cin >> H >> W;
  for(int i=0;i<H;i++){
    for(int j=0;j<W;j++){
      cin >> t;
      maze[i][j] = (t=='.');
      dist[i][j] = -1*(t=='.');
      if(t=='#') que.push(make_pair(i,j));
    }
  }
  while(!que.empty()){
    int cy=que.front().first,cx=que.front().second;
    que.pop();
    ans=dist[cy][cx];
    for(int i=0;i<4;i++){
      int ny=cy+dy[i],nx=cx+dx[i];
      if(ny<0||H<=ny||nx<0||W<=nx||maze[ny][nx]=='#')
        continue;
      if(dist[ny][nx]==-1){
        dist[ny][nx]=dist[cy][cx]+1;
        que.push(make_pair(ny,nx));
      }
    }
  }
  cout << ans << "\n";
}
