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;


struct P
{
    int x,y;
};
int h,w,k;
char a[1111][1111];
int o[1111][1111],dx[10]={1,0,-1,0},dy[10]={0,1,0,-1};
queue<P> q;


int main()
{
    memset(o,-1,sizeof(o));
	scanf("%d %d",&h,&w);
	for(int i=1;i<=h;i++)
        scanf("%s",a[i]+1);
    for(int i=1;i<=h;i++)
        for(int j=1;j<=w;j++)
        if(a[i][j]=='#')
    {
        o[i][j]=0;
        q.push({i,j});
    }
    for(;q.size();q.pop())
    for(int h=0;h<4;h++)
    {
        int x=dx[h]+q.front().x,y=dy[h]+q.front().y;
        if(o[x][y]<0&&a[x][y])
        {
            o[x][y]=o[q.front().x][q.front().y]+1;
            k=max(k,o[x][y]);
            q.push({x,y});
        }
    }
    printf("%d",k);
}
