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

struct st {int r, c;};
int h, w, cnt, ans, v[1005][1005];
char a[1005][1005];
queue<st> qu;


void push(int p, int q, int r) {
	if(p>=0 && p<h && q>=0 && q<w && v[p][q]==-1) {
		qu.push({p, q});
		v[p][q] = r;
	}
}

int main() {
	int i, j;
	memset(v, -1, sizeof(v));
	cin >> h >> w;
	for(i=0; i<h; i++) {
		scanf("%s", a[i]);
		for(j=0; j<w; j++) {
			if(a[i][j]=='#')
				push(i, j, 0);
		}
	}
	while(!qu.empty()) {
		st t = qu.front(); qu.pop();
		cnt++;
		if(cnt == w*h) {
			cout << v[t.r][t.c];
			break;
		}
		push(t.r-1, t.c, v[t.r][t.c]+1);
		push(t.r+1, t.c, v[t.r][t.c]+1);
		push(t.r, t.c-1, v[t.r][t.c]+1);
		push(t.r, t.c+1, v[t.r][t.c]+1);
	}
//	for(i=0; i<h; i++, puts("")) for(j=0; j<w; j++) printf("%d", v[i][j]);
	return 0;
}