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 h, w;
	cin >> h >> w;
	string a[1111];
	for (int i = 0; i < h; i++)
		cin >> a[i];
	queue<tuple<int, int, int>> q;
	for (int i = 0; i < h; i++)
		for (int j = 0; j < w; j++)
			if (a[i][j] == '#')
				q.emplace(i, j, 0);
	int r = 0, s[] = { 0,1,0,-1,0 };
	while (!q.empty()) {
		auto t = q.front();
		q.pop();
		r = max(r, get<2>(t));
		for (int i = 0; i < 4; i++) {
			int x = get<0>(t) + s[i], y = get<1>(t) + s[i + 1];
			if (0 <= x && x < h && 0 <= y && y < w&&a[x][y] == '.')
				q.emplace(x, y, get<2>(t) + 1), a[x][y] = '#';
		}
	}
	cout << r << endl;
	return 0;
}