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<queue>
#include<algorithm>
using namespace std;

char c[1001][1002];
int d[1000001];
int main(){
	int n,m;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
		scanf("%s",c[i]+1);
	queue<int>q;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			if(c[i][j]=='#'){
				q.push((i-1)*m+j);
				d[(i-1)*m+j]=1;
			}
	int ans=0;
	while(!q.empty()){
		int u=q.front();
		q.pop();
		ans=max(ans,d[u]);
		for(int i=0;i<4;i++){
			int v=u;
			if(i==0&&u<=m||i==1&&u%m==1%m||i==2&&u>=(n-1)*m+1||i==3&&u%m==0)
				continue;
			if(i==0)v-=m;
			else if(i==1)v--;
			else if(i==2)v+=m;
			else v++;
			if(!d[v]){
				d[v]=d[u]+1;
				q.push(v);
			}
		}
	}
	printf("%d\n",ans-1);
}