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>
#define ll          long long
using namespace std;
string s[1005];
int vis[1005][1005];
int n,m;
int fx[]={1,-1,0,0};
int fy[]={0,0,-1,1};
int chk(int x,int y){
	if(vis[x][y])return 0;
	if(x<0||y<0||x>n-1||y>m-1)return 0;
	if(s[x][y]=='#')return 0;
	return 1;	
}
void bfs(){
	queue<pair<int,int>> q;
	for(int i=0;i<n;i++){
		for(int j=0;j<m;j++){
			if(s[i][j]=='#'){
				q.push({i,j});
				vis[i][j]=1;
			}
		}
	}
	while(!q.empty()){
		pair<int,int> p=q.front();
		q.pop();
		int x=p.first,y=p.second;
		for(int i=0;i<4;i++){
			if(chk(fx[i]+x,fy[i]+y)){
				q.push({fx[i]+x,fy[i]+y});
				vis[fx[i]+x][fy[i]+y]=vis[x][y]+1;
			}
		}
	}
}
int main(){
	cin>>n>>m;
	for(int i=0;i<n;i++)cin>>s[i];
	bfs();
	int mx=0;
	for(int i=0;i<n;i++){
		for(int j=0;j<m;j++){
			mx=max(mx,vis[i][j]);
		}
	}
	cout<<mx-1<<endl;
}