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<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define rep(i,l,r) for (int i=(l); i<=(r); i++)
typedef long long ll;
using namespace std;

const int N=2010,inf=1e8;
char mp[N][N];
int n,m,ans,f[N][N];

int main(){
	scanf("%d%d",&n,&m);
	rep(i,1,n) rep(j,1,m) scanf(" %c",&mp[i][j]);
	rep(i,1,n) rep(j,1,m){
		if (mp[i][j]=='#') f[i][j]=0; else f[i][j]=inf;
		if (i>1) f[i][j]=min(f[i][j],f[i-1][j]+1);
		if (j>1) f[i][j]=min(f[i][j],f[i][j-1]+1);
	}
	for (int i=n; i; i--) for (int j=m; j; j--){
		if (i<n) f[i][j]=min(f[i][j],f[i+1][j]+1);
		if (j<m) f[i][j]=min(f[i][j],f[i][j+1]+1);
	}
	rep(i,1,n) rep(j,1,m) ans=max(ans,f[i][j]);
	printf("%d\n",ans);
	return 0;
}