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;
int h,w;
int dx[]={0,0,1,-1};
int dy[]={1,-1,0,0};
char s[1005][1005];
struct node
{
    int x,y,res;
    node(int x=0,int y=0,int res=0):x(x),y(y),res(res){}
};
queue<node>q;
int main()
{
    scanf("%d%d",&h,&w);
    for(int i=1;i<=h;i++)
        scanf("%s",s[i]+1);
    for(int i=1;i<=h;i++)
        for(int j=1;j<=w;j++)
    {
        if(s[i][j]=='#')
            q.push(node(i,j,0));
    }
    int ans=0;
    while(!q.empty())
    {
        node now=q.front();q.pop();
        ans=max(ans,now.res);
        for(int i=0;i<4;i++)
        {
            int xx=dx[i]+now.x;
            int yy=dy[i]+now.y;
            if(xx<1||xx>h||yy<1||yy>w||s[xx][yy]=='#') continue;
            s[xx][yy]='#';
            q.push(node(xx,yy,now.res+1));
        }
    }
    cout<<ans<<endl;
}
