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;
using ll = long long;

int main()
{
	ll h,w;
	cin >> h >> w;
	vector<string> s(h);
	for(auto &i:s)cin >> i;
	vector<vector<ll>> dis(h,vector<ll>(w,1e9));

	for(int i = 0;i<h;i++)
	{
		for(int j = 0;j<w;j++)
		{
			if(s[i][j]=='#')dis[i][j]=0;
		}
	}

	for(int i = 0;i<h;i++)
	{
		for(int j = 1;j<w;j++)
		{
			dis[i][j] = min(dis[i][j],dis[i][j-1]+1);
		}
	}

	for(int i = 0;i<h;i++)
	{
		for(int j = w-2;j>=0;j--)
		{
			dis[i][j] = min(dis[i][j],dis[i][j+1]+1);
		}
	}
	for(int i = 1;i<h;i++)
	{
		for(int j = 0;j<w;j++)
		{
			dis[i][j] = min(dis[i][j],dis[i-1][j]+1);
		}
	}

	for(int i = h-2;i>=0;i--)
	{
		for(int j = 0;j<w;j++)
		{
			dis[i][j] = min(dis[i][j],dis[i+1][j]+1);
		}
	}
	ll ans = 0;
	for(int i = 0;i<h;i++)for(int j = 0;j<w;j++)ans = max(ans,dis[i][j]);
	cout << ans << endl;

	
}