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;

typedef pair<int, int> pii;
const int N = 1010;

int n, m, dis[N][N];
int delta_x[] = {0, 0, 1, -1},
	delta_y[] = {1, -1, 0, 0};
char a[N][N];
queue<pii> q;

int main() {
	ios::sync_with_stdio(false), cin.tie(0);
	cin >> n >> m;
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++) {
			cin >> a[i][j];
			if(a[i][j] == '#')
				q.push(pii(i, j));
		}
	while(q.size()) {
		pii p = q.front();
		q.pop();
		int x = p.first, y = p.second;
		for (int i = 0; i < 4; i++) {
			int X = x + delta_x[i],
				Y = y + delta_y[i];
			if(a[X][Y] == '.' && dis[X][Y] == 0) {
				dis[X][Y] = dis[x][y] + 1;
				q.push(pii(X, Y));
			}
		}
	}
	int ans = 0;
	for (int i = 1; i <= n; i++)
		ans = max(ans, *max_element(dis[i], dis[i] + N));
	cout << ans;
	return 0;
}