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;
#define mp(a, b) make_pair(a, b)
typedef pair<int, int> pii; 
const int maxn = 1000 + 10;
int n, m;
const int dx[] = {0, -1, 0, 1};
const int dy[] = {-1, 0, 1, 0};
queue<pii> Q;
char s[maxn][maxn];
int vis[maxn][maxn], ans;
int main()
{
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; ++i)
		scanf("%s", s[i] + 1);
	for(int i = 1; i <= n; ++i)
		for(int j = 1; j <= m; ++j)
		{
			vis[i][j] = -1;
			if(s[i][j] == '#')
			{ 
				vis[i][j] = 0; 
				Q.push(mp(i, j));
			}
		}
	while(!Q.empty())
	{
		pii x = Q.front(); Q.pop();
		ans = max(ans, vis[x.first][x.second]);
		for(int i = 0; i < 4; ++i)
			if(vis[x.first + dx[i]][x.second + dy[i]] == -1)
			{
				vis[x.first + dx[i]][x.second + dy[i]] = ans + 1;
				Q.push(mp(x.first + dx[i], x.second + dy[i]));
			}
	}
	printf("%d\n", ans);
	return 0;
}
	
