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 r(i,n) for(int i=0;i<n;i++)
using namespace std;
#define int long long
typedef pair<int,int>P;

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

int dp[1111][1111],h,w,ans;

string s[1111];
signed main(){
  memset(dp,-1,sizeof(dp));
  cin>>h>>w;
  r(i,h)cin>>s[i];
  queue<P>q;
  r(i,h)r(j,w){
    if(s[i][j]=='#'){
      q.push(P(i,j));
      dp[i][j]=0;
    }
  }
  while(!q.empty()){
    P p=q.front();q.pop();
    int y=p.first;
    int x=p.second;
    ans=max(ans,dp[y][x]);
    r(i,4){
      int Y=y+dy[i];
      int X=x+dx[i];
      if(Y<0||X<0||h<=Y||w<=X)continue;
      if(dp[Y][X]!=-1)continue;
      dp[Y][X]=dp[y][x]+1;
      q.push(P(Y,X));
    }
  }
  cout<<ans<<endl;
}
