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<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<cmath>
#include<vector>
#include<queue>
using namespace std;
const int N=1e3+5;
char s[N][N];
int dy[]={1,0,-1,0};
int dx[]={0,1,0,-1};
struct node{
    int y,x,s;
};
int ans;
queue<node>q;
int main(){
    int n,m;scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++) scanf("%s",s[i]+1);
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++)if(s[i][j]=='#'){
            q.push((node){i,j,0});
        }
    }
    while(!q.empty()){
        node t=q.front();q.pop();
        ans=max(ans,t.s);
        for(int i=0;i<4;i++){
            int yy=dy[i]+t.y;
            int xx=dx[i]+t.x;
            if(yy<1||xx<1||yy>n||xx>m||s[yy][xx]=='#') continue;
            s[yy][xx]='#';
            q.push((node){yy,xx,t.s+1});
        }
    }
    printf("%d",ans);
}
