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>
using namespace std;
int n, m;
char s[1010][1010];
queue <pair<int,int> >q;
int st[1010][1010], vis[1010][1010];
const int X[]={0,0,1,-1};
const int Y[]={1,-1,0,0};
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1; i<=n; ++i){
		scanf("%s",s[i]);
		for(int j=1; j<=m; ++j)if(s[i][j-1]=='#')q.push(make_pair(i,j)),vis[i][j]=1;
	}
	int mxst=0;
	while(!q.empty()){
		int x=q.front().first, y=q.front().second; q.pop();
		mxst=max(mxst,st[x][y]);
		for(int i=0; i<4; ++i){
			int dx=x+X[i],dy=y+Y[i];
			if(dx>0&&dy>0&&dx<=n&&dy<=m&&!vis[dx][dy]){
				q.push(make_pair(dx,dy)); st[dx][dy]=st[x][y]+1; vis[dx][dy]=1;
			}
		}
	}
	printf("%d",mxst);
	return 0;
}