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 n,m,a[1005][1005],x,y,curx,cury,maks=0;
int gx[4]={1,-1,0,0};
int gy[4]={0,0,1,-1};
char c;
queue<pair<int,int> > q;
int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL); cout.tie(NULL);
	cin>>n>>m;
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++){
			cin>>c;
			if(c=='#'){
				a[i][j]=0;
				q.push(make_pair(i,j));
			}
			else{
				a[i][j]=-1;
			}
		}
	}
	while(!q.empty()){
		x=q.front().first;
		y=q.front().second;
		q.pop();
		for(int i=0;i<4;i++){
			curx=x+gx[i];cury=y+gy[i];
			if(a[curx][cury]==-1){
				a[curx][cury]=a[x][y]+1;
				if(maks<a[curx][cury]) maks=a[curx][cury];
				q.push(make_pair(curx,cury));
			}
		}
	}
	cout<<maks<<endl;
}