Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
typedef pair<int,int> P;

int h,w; char f[1000][1000];
int dx[4]={-1,0,1,0},dy[4]={0,1,0,-1};

int bfs(int n,queue<P> que){
    int l=que.size();
    if(l==0)return n-1;
    for (int i = 0; i < l; i++){
        P p=que.front(); que.pop();
        for (int j = 0; j < 4; j++){
            int ny=p.first+dy[j],nx=p.second+dx[j];
            if(ny>=0&&ny<h&&nx>=0&&nx<w&&f[ny][nx]=='.'){
                f[ny][nx]='#'; que.push(P(ny,nx));
            }
        }
    }
    return bfs(n+1,que);
}

int main(){
    cin.tie(0); ios::sync_with_stdio(false);
    
    cin>>h>>w; queue<P> que;
    for (int i = 0; i < h; i++){
        for (int j = 0; j < w; j++){
            char c; cin>>f[i][j];
            if(f[i][j]=='#')que.push(P(i,j));
        }
    }
    cout<<bfs(0,que)<<endl;
}