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>
#define int long long
using namespace std;
int dx[]={0,1,0,-1};
int dy[]={1,0,-1,0};
signed main(){
  int H,W;
  cin>>H>>W;
  queue<pair<int,int>> Q;
  vector<vector<char>> A(H,vector<char>(W));
  vector<vector<int>> B(H,vector<int>(W,-1));
  for(int i=0;i<H;i++)
    for(int j=0;j<W;j++){
      cin>>A[i][j];
      if(A[i][j]=='#'){
        Q.push({i,j});
        B[i][j]=0;
      }
    }
  int ans=0;
  while(!Q.empty()){
    pair<int,int> p=Q.front();
    Q.pop();
    for(int i=0;i<4;i++){
      int x=p.first+dx[i],y=p.second+dy[i];
      if(0<=x && x<H && 0<=y && y<W && B[x][y]==-1){
        B[x][y]=B[p.first][p.second]+1;
        Q.push({x,y});
        ans=max(ans,B[x][y]);
      }
    }
  }
  cout<<ans<<endl;
}