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 main() {
	int n, m;
	cin >> n >> m;
	vector<int> vis(n * m);
	queue<int> q;
	for (int i = 0; i < n; i++) {
		string str;
		cin >> str;
		for (int j = 0; j < m; j++) {
			int now = i * m + j;
			if (str[j] == '#') {
				vis[now] = 1;
				q.push(now);
			}
		}
	}
	int ans = 0;
	while (q.size()) {
		queue<int> nq;
		while (q.size()) {
			int u = q.front();
			q.pop();
			int x = u / m;
			int y = u % m;
			if (x && !vis[u - m]) {
				vis[u - m] = 1;
				nq.push(u - m);
			}
			if (y && !vis[u - 1]) {
				vis[u - 1] = 1;
				nq.push(u - 1);
			}
			if (x < n - 1 && !vis[u + m]) {
				vis[u + m] = 1;
				nq.push(u + m);
			}
			if (y < m - 1 && !vis[u + 1]) {
				vis[u + 1] = 1;
				nq.push(u + 1);
			}
		}
		if (nq.size()) {
			ans++;
		}
		q = nq;
	}
	cout << ans << endl;
	return 0;
}