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 <queue>

using namespace std;

const int MAXN = 1e3 + 5;
const int X[] = {1, 0, -1, 0};
const int Y[] = {0, 1, 0, -1};

int m, n, ans = 0;
char ch[MAXN][MAXN];
bool vis[MAXN][MAXN];

struct Data{
	int x, y, val;
};

void Bfs() {
	queue<Data> q;
	for (int i = 1; i <= m; i++) {
		for (int j = 1; j <= n; j++) {
			if (ch[i][j] == '#') {
				q.push((Data){i, j, 0});
				vis[i][j] = true;
			}
		}
	}
	while (!q.empty()) {
		Data now = q.front(); q.pop();
		ans = max(ans, now.val);
		for (int i = 0; i < 4; i++) {
			int nx = X[i] + now.x, ny = Y[i] + now.y;
			if (nx < 1 || nx > m || ny < 1 || ny > n) continue;
			if (ch[nx][ny] != '#') q.push((Data){nx, ny, now.val + 1}), ch[nx][ny] = '#';
		}
	}
}

int main() {
	cin >> m >> n;
	for (int i = 1; i <= m; i++) cin >> ch[i] + 1;
	Bfs();
	cout << ans << endl;
	return 0;
}