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 <iostream>
#include <vector>
#include <set>
using namespace std;

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);

	int h, w;
	cin >> h >> w;
	vector<vector<char>> a(h);
	set<pair<int, int>> p;
	for (int i = 0; i < h; i++) {
		a[i].resize(w);
		for (int j = 0; j < w; j++) {
			cin >> a[i][j];
			if (a[i][j] == '#')
				p.insert({ i, j });
		}
	}
	int d[][2] = { -1, 0, 1, 0, 0, -1, 0, 1 };
	int ans = 0;
	while (!p.empty()) {
		set<pair<int, int>> s;
		for (auto const& x : p) {
			for (auto const y : d) {
				int i = x.first + y[0], j = x.second + y[1];
				if (i >= 0 && i < h && j >= 0 && j < w && a[i][j] == '.')
					s.insert({ i, j});
			}
		}
		if (!s.empty()) {
			ans++;
			for (auto const& x : s)
				a[x.first][x.second] = '#';
		}
		p = move(s);
	}
	cout << ans << '\n';

	return 0;
}
