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;
const int maxn=1e3+5;
const int dx[]={-1,0,1,0},dy[]={0,-1,0,1};
char a[maxn][maxn];
int v[maxn][maxn];
int n,m;
int ans;
struct Node{
	int x;
	int y;
	int d;
};
queue<Node>q;
int main(){	
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++) scanf("%s",a[i]+1);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			if(a[i][j]=='#'){
				v[i][j]=1;
				q.push((Node){i,j,0});
			}
	while(!q.empty()){
		int x=q.front().x;
		int y=q.front().y;
		int d=q.front().d;
		ans=d;
		q.pop();
		for(int i=0;i<4;i++){
			int xx=x+dx[i],yy=y+dy[i];
			if(xx<1||xx>n||yy<1||yy>m) continue;
			if(v[xx][yy]) continue;
			if(a[xx][yy]=='#') continue;
			v[xx][yy]=1;
			q.push((Node){xx,yy,d+1});
		}
	}
	printf("%d\n",ans);
	return 0;
}