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;

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

char s[1100][1100];
int n, m;
struct task {
	int x, y;
	int res;
};
queue<task>q;


int main() {
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; ++i) {
		scanf("%s", s[i] + 1);
	}
	for(int i = 1; i <= n; ++i) {
		for(int j = 1; j <= m; ++j) {
			if(s[i][j] == '#') q.push({i, j, 0});
		}
	}
	int ans = 0;
	while(!q.empty()) {
		task now = q.front(); q.pop();
		ans = max(ans, now.res);
		for(int i = 0; i < 4; ++i) {
			int x = now.x + dx[i], y = now.y + dy[i];
			if(s[x][y] == '.') {
				s[x][y] = '#';
				q.push({x, y, now.res + 1});
			}
		}
	}
	cout << ans << endl;
}