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 mp(x,y) make_pair(x,y)
#define pa pair<int,int>
using namespace std;
const int N=1010;
const int dx[4]={0,0,1,-1};
const int dy[4]={1,-1,0,0};
int n,m;
queue<pa>q;
char s[N][N];
int ans,b[N][N];
void dfs()
{
	while (!q.empty())
	{
		pa t=q.front(); q.pop();
		for (int k=0;k<4;k++) {
			int i=t.first+dx[k];
			int j=t.second+dy[k];
			if (i>0&&i<=n&&j>0&&j<=m) {
				if (s[i][j]=='.') {
					s[i][j]='#';
					q.push(mp(i,j)); b[i][j]=b[t.first][t.second]+1;
					ans=max(ans,b[i][j]);
				}
			}
		}	
	}
}
int main()
{
	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(mp(i,j));
		}
	}
	dfs();
	printf("%d\n",ans);
}