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 dp[1005][1005],n,m,ans;
int min(int x,int y,int z,int w){
	int ans=x;
	if (y<ans) ans=y;
	if (z<ans) ans=z;
	if (w<ans) ans=w;
	return ans;
}
int main(){
	char ch;
	cin>>n>>m;
	
	for (int i=0;i<=n+1;i++){
		for (int j=0;j<=m+1;j++){
			dp[i][j]=1000000;
		}
	}
	for (int i=1;i<=n;i++){
		for (int j=1;j<=m;j++){
			cin>>ch;
			if (ch=='#') dp[i][j]=0;
		}
	}
	
	for (int i=1;i<=n;i++){
		for (int j=1;j<=m;j++){
			if (dp[i][j]!=0) dp[i][j]=min(dp[i-1][j],dp[i+1][j],dp[i][j-1],dp[i][j+1])+1;
		}
	}
	for (int i=n;i>=1;i--){
		for (int j=m;j>=1;j--){
			if (dp[i][j]!=0) dp[i][j]=min(dp[i-1][j],dp[i+1][j],dp[i][j-1],dp[i][j+1])+1;
		}
	}
	
	for (int i=1;i<=n;i++){
		for (int j=1;j<=m;j++){
			ans=max(ans,dp[i][j]);
		}
	}
	cout<<ans;
	return 0;
}