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 h, w;
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
const int N = 1010;
char a[N][N];
int d[N][N], vis[N][N];
int main() {
	scanf("%d%d", &h, &w);
	for (int i = 1; i <= h; ++i)
		scanf("%s", a[i] + 1);
	queue<pair<int, int> > q;
	for (int i = 1; i <= h; ++i)
		for (int j = 1; j <= w; ++j)
			if (a[i][j] == '#') vis[i][j] = 1,q.push(make_pair(i, j));
	int ans = 0;
	while (!q.empty()) {
		int x = q.front().first, y = q.front().second;
		q.pop();
		ans = d[x][y];
		for (int i = 0; i < 4; ++i) {
			int nx = x + dx[i], ny = y + dy[i];
			if (!vis[nx][ny] && a[nx][ny] == '.') {
				vis[nx][ny] = 1;
				d[nx][ny] = d[x][y] + 1;
				q.push(make_pair(nx, ny));
			}
		}
	}
	printf("%d", ans);
	return 0;
}