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
#include <bits/stdc++.h>
using namespace std;

int main() {
	// your code goes here
	int n,m;
	cin>>n>>m;
	char mat[n][m];
	queue<pair<int,pair<int,int>>>q;
	for(int i=0;i<n;i++){
		for(int j=0;j<m;j++){
			cin>>mat[i][j];
			if(mat[i][j]=='#')
				q.push(make_pair(0,make_pair(i,j)));
		}
	}
	int x[]={0,0,1,-1};
	int y[]={-1,1,0,0};
	int ans=0;
	while(!q.empty()){
		pair<int,pair<int,int>>p=q.front();
		q.pop();
		ans=max(ans,p.first);
		for(int i=0;i<4;i++){
			int r=p.second.first+x[i];
			int s=p.second.second+y[i];
			if(r>=0 && r<n && s>=0 && s<m && mat[r][s]=='.'){
				mat[r][s]='#';
				q.push(make_pair(p.first+1,make_pair(r,s)));
			}
		}
	}
	cout<<ans;
	return 0;
}