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>
#define F first
#define S second
using namespace std;

int h, w, r, c, v[1005][1005];
char s[1005];
queue<pair<int, int>> qu;

void push(int p, int q, int r) {
	if(!v[p][q]) {
		v[p][q] = r;
		qu.push({p, q});
	}
}

int main() {
	int i, j;
	memset(v, -1, sizeof(v));
	cin >> h >> w;
	for(i=1; i<=h; i++) {
		scanf("%s", &s[1]);
		for(j=1; j<=w; j++) {
			v[i][j] = 0;
			if(s[j]=='#') {
				v[i][j] = 1;
				qu.push({i, j});
			}
		}
	}
	while(!qu.empty()) {
		r = qu.front().F;
		c = qu.front().S;
		qu.pop();
		push(r+1, c, v[r][c]+1);
		push(r-1, c, v[r][c]+1);
		push(r, c+1, v[r][c]+1);
		push(r, c-1, v[r][c]+1);
	}
	cout << v[r][c] - 1;
	return 0;
}
