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 ll long long
using namespace std;
const ll INF=(ll)1e9;
const ll MOD=(ll)1e9+7;
vector<int> dx={1,0,-1,0},dy={0,1,0,-1};

int main(){
  int H,W,cnt=0,nx,ny;
  queue<pair<int,int>> q;
  pair<int,int> p;
  cin>>H>>W;
  vector<vector<int>> A(H,vector<int>(W));
  for(int i=0;i<H;i++){
    for(int j=0;j<W;j++){
      char cha;
      cin>>cha;
      if(cha=='#'){
        A[i][j]=0;
        q.push(pair<int,int>(i,j));
      }
      else A[i][j]=INF;
    }
  }
  while(q.size()){
    pair<int,int> P=q.front();
    q.pop();
    for(int k=0;k<4;k++){
      nx=P.second+dx[k],ny=P.first+dy[k];
      if(0<=nx&&nx<W&&0<=ny&&ny<H&&A[ny][nx]==INF){
        A[ny][nx]=A[P.first][P.second]+1;
        cnt=A[ny][nx];
        q.push(pair<int,int>(ny,nx));
      }
    }
  }
  cout<<cnt<<endl;
}
