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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

#define ran 1011

const int go[4][2] = {0, 1, 1, 0, 0, -1, -1, 0};

int n, m;
char s[ran][ran];
int d[ran][ran];
int q[ran*ran*2], l, r;

int main() {
	scanf("%d%d", &n, &m);
	memset(d, -1, sizeof d);
	l = r = 0;
	for(int i=0;i<n;i++) {
		scanf("%s", s[i]);
		for(int j=0;j<m;j++)
			if(s[i][j]=='#') {
				q[r++] = i;
				q[r++] = j;
				d[i][j] = 0;
			}
	}
	
	int ans=0;
	while(l<r) {
		int cur_x = q[l++], cur_y = q[l++];
		int cur_d = d[cur_x][cur_y];
		for(int i=0;i<4;i++) {
			int x = cur_x + go[i][0], y = cur_y + go[i][1];
			if(x<0 || x>=n || y<0 || y>=m) continue;
			if(d[x][y]>=0) continue;
			
			d[x][y] = cur_d + 1;
			q[r++] = x; q[r++] = y;
			ans = max(ans, d[x][y]);
		}
	}
	printf("%d\n", ans);
	
	return 0;
}