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>
#define MAX 1000
#define mp make_pair
using namespace std;
typedef pair<int,int> pii;
int n,m,a[MAX+5][MAX+5];
int dir1[4]={0,0,-1,1},dir2[4]={1,-1,0,0};
queue<pii>q;
int out(int x,int y)
{
	return x<1||y<1||x>n||y>m;
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1; i<=n; i++)
	{
	cin.ignore();
    for(int j=1; j<=m; j++)
	{
	scanf("%c",&a[i][j]);
	if(a[i][j]=='#') q.push(mp(i,j)); 
    }
    }
    int sz=q.size(),cnt=0;
    while(!q.empty())
    {
    	int x=q.front().first,y=q.front().second; q.pop();
    	if(!sz) cnt++,sz=q.size();
    	else sz--;
    	for(int i=0; i<4; i++)
    	{
    		int xs=x+dir1[i],ys=y+dir2[i];
    		if(out(xs,ys)||a[xs][ys]=='#') continue;
			q.push(mp(xs,ys)); a[xs][ys]='#';
    	}
    }
    printf("%d",cnt);
	return 0;
}