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 mx[ ]={0,1,0,-1};
int my[ ]={1,0,-1,0};

int main(){
	queue<pair<pair<int,int>,int> > bfs;
	int h,w;
	cin >> h >> w;
	char area[h+5][w+5];
	bool vis[h+5][w+5];
	memset(vis,0,sizeof vis);
	for(int i=0;i<h;i++){
		for(int j=0;j<w;j++){
			cin >> area[i][j];
			if(area[i][j]=='#'){
				bfs.push({{i,j},0});
				vis[i][j]=1;
			}
		}
	}
	int ans=0;
	while(!bfs.empty()){
		int x=bfs.front().first.first;
		int y=bfs.front().first.second;
		int n=bfs.front().second;
		ans=max(ans,n);
		bfs.pop();
		for(int i=0;i<4;i++){
			if(x+mx[i]>=0 && y+my[i]>=0 && x+mx[i]<h && y+my[i]<w && !vis[x+mx[i]][y+my[i]]){
				bfs.push({{x+mx[i],y+my[i]},n+1});
				vis[x+mx[i]][y+my[i]]=1;
			}
		}
	}
	cout << ans << '\n';
}