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;

int main() {
	int H, W;
	cin >> H >> W;

	vector<string> A(H);
	vector<vector<int>> d(H, vector<int>(W, 1e9));
	queue<pair<int,int>> q;
	for (int i = 0; i < H; ++i) {
		cin >> A[i];
		for (int j = 0; j < W; ++j) {
			if (A[i][j] == '#') {
				d[i][j] = 0;
				q.emplace(j, i);
			}
		}
	}

	int dx[]{1, -1, 0, 0}, dy[]{0, 0, 1, -1};
	int ans = 0;
	while (!q.empty()) {
		auto p = q.front(); q.pop();
		int x = p.first, y = p.second;
		for (int i = 0; i < 4; ++i) {
			int nx = x + dx[i], ny = y + dy[i];
			if (nx < 0 || W <= nx || ny < 0 || H <= ny || A[ny][nx] == '#') {
				continue;
			}
			A[ny][nx] = '#';
			q.emplace(nx, ny);
			ans = max(d[ny][nx] = d[y][x] + 1, ans);
		}
	}

	cout << ans << endl;

	return 0;
}