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;

const int dx[] = { 0, 1, 0, -1 };
const int dy[] = { 1, 0, -1, 0 };

int main()
{
	int H, W;
	cin >> H >> W;
	vector<string> S(H);
	vector<pair<int, int>> pos;
	for (int i = 0; i < H; i++) {
		cin >> S[i];
		for (int j = 0; j < W; j++) {
			if (S[i][j] == '#') {
				pos.emplace_back(i, j);
			}
		}
	}
	int cnt = -1;
	while (!pos.empty()) {
		vector<pair<int, int>> tmp;
		for (auto p : pos) {
			int x, y; tie(x, y) = p;
			for (int i = 0; i < 4; i++) {
				int tx = x + dx[i], ty = y + dy[i];
				if (0 <= tx && tx < H && 0 <= ty && ty < W && S[tx][ty] != '#') {
					S[tx][ty] = '#';
					tmp.emplace_back(tx, ty);
				}
			}
		}
		pos = move(tmp);
		++cnt;
	}
	cout << cnt << endl;
	return 0;
}