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;
const int N=1005;
char base[N][N];
int dist[N][N];
int Move[4][2]={{1,0},{0,1},{-1,0},{0,-1}};
int main()
{
	int n,m;
	scanf("%i %i",&n,&m);
	for(int i=1;i<=n;i++) scanf("%s",base[i]+1);
	queue<int> q;
	for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) if(base[i][j]=='#') q.push(i*N+j);
	int ans;
	while(q.size())
	{
		int x=q.front()/N;
		int y=q.front()%N;
		q.pop();
		ans=dist[x][y];
		for(int i=0;i<4;i++)
		{
			int nx=x+Move[i][0];
			int ny=y+Move[i][1];
			if(base[nx][ny]=='.' && !dist[nx][ny]) dist[nx][ny]=dist[x][y]+1,q.push(nx*N+ny);
		}
	}
	printf("%i\n",ans);
	return 0;
}