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;
#define rep(i,a) for(int i=0;i<(a);i++)
const ll MOD=1000000007;
// const ll MOD=998244353;

int A[1010][1010];
struct za{
  int x;
  int y;
  int num;
};

int dx[4]={1,-1,0,0};
int dy[4]={0,0,1,-1};

int main(){
  int H,W; cin>>H>>W;

  queue<za> que;
  rep(i,H) rep(j,W){
    char c; cin>>c;
    if(c=='#'){
      A[i][j]=1;
      que.push({i,j,0});
    }
  }

  int ans=0;
  while(que.size()){
    struct za z=que.front();
    que.pop();
    rep(k,4){
      int i=z.x+dx[k],j=z.y+dy[k];
      if(i<0||i>=H||j<0||j>=W) continue;
      if(A[i][j]==0){
        A[i][j]=1;
        que.push({i,j,z.num+1});
        ans=max(ans,z.num+1);
      }
    }
  }
  cout<<ans<<endl;
  return 0;
}
