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 dirx[4] = {1, 0, -1, 0};
const int diry[4] = {0, 1, 0, -1};

char s[1005][1005];
int used[1005][1005];
queue < tuple <int, int, int> > q;
int h, w;

bool check(int x, int y){
	return x >= 1 && x <= h && y >= 1 && y <= w;
}

int main(){
	int ans, x, y, d, x0, y0;
	scanf("%d %d",&h,&w);
	for (int i = 1; i <= h; i++) scanf("%s", s[i] + 1);
	ans = 0;
	for (int i = 1; i <= h; i++){
		for (int j = 1; j <= w; j++){
			if (s[i][j] == '#'){
				q.push(make_tuple(i, j, 0));
				used[i][j] = 1;
			}
		}
	}
	while (!q.empty()){
		tie(x, y, d) = q.front(); q.pop();
		ans = max(ans, d);
		for (int i = 0; i < 4; i++){
			x0 = x + dirx[i]; y0 = y + diry[i];
			if (check(x0, y0) && !used[x0][y0]){
				q.push(make_tuple(x0, y0, d + 1));
				used[x0][y0] = 1;
			}
		}
	}
	printf("%d\n", ans);
	return 0;
}