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;

const int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};

int main(int argc, char const *argv[])
{
	int h,w;
	cin>>h>>w;
	int memo[h][w];
	char a;
	queue<pair<int,int>> que;
	for (int i = 0; i < h; ++i)
	{
		for (int j = 0; j < w; ++j)
		{
			cin>>a;
			if (a=='.')
			{
				memo[i][j]=-1;
			}
			else
			{
				memo[i][j]=0;
				que.push(make_pair(i,j));
			}
		}
	}
	int ans=0;
	while (que.size()>0)
	{
		auto now=que.front();que.pop();
		for (int i = 0; i < 4; ++i)
		{
			int nx=now.first+dx[i],ny=now.second+dy[i];
			if (nx<0||h<=nx||ny<0||w<=ny)
			{
				continue;
			}
			if (memo[nx][ny]==-1)
			{
				memo[nx][ny]=memo[now.first][now.second]+1;
				ans=max(ans,memo[nx][ny]);
				que.push(make_pair(nx,ny));
			}
		}
	}
	cout<<ans<<endl;
	return 0;
}
