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;

char c[1010][1010];
queue < pair<pair<int,int>,int> > Q;
const int dx[4]={0,0,1,-1};
const int dy[4]={1,-1,0,0};
bool vis[1010][1010];
int n,m;

bool check(int x,int y){
	return x>=0&&x<n&&y>=0&&y<m;
}

int main(){
	scanf("%d%d",&n,&m);
	for(int i=0;i<n;i++){
		for(int j=0;j<m;j++){
			cin>>c[i][j];
			if(c[i][j]=='#'){
				vis[i][j]=1;
				Q.push(make_pair(make_pair(i,j),0));
			}
		}
	}
	int ans=0;
	while(!Q.empty()){
		int x=Q.front().first.first;
		int y=Q.front().first.second;
		int d=Q.front().second;
		ans=max(ans,d);
		Q.pop();
		for(int i=0;i<4;i++){
			int nx=x+dx[i];
			int ny=y+dy[i];
			if(check(nx,ny)){
				if(!vis[nx][ny]&&c[nx][ny]=='.'){
					vis[nx][ny]=1;
					Q.push(make_pair(make_pair(nx,ny),d+1));
				}
			}
		}
	}
	printf("%d\n",ans);
	return 0;
}