Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m, D[1010][1010], dx[4] = { 0,1,0,-1 }, dy[4] = { 1,0,-1,0 };
queue <pair<int,int> > q;
char p[1010][1010];
int ans = -1;
void Put(int x, int y, int d) {
	if (x<1 || x>n || y<1 || y>m)return;
	if (D[x][y] <= d)return;
	D[x][y] = d;
	ans = max(ans,D[x][y]);
	q.push({x,y});
}
int main() {
	int i, j;
	scanf("%d%d", &n, &m);
	for (i = 1; i <= n; i++) {
		scanf("%s", p[i] + 1);
		for (j = 1; j <= m; j++) {
			D[i][j] = 1e9;
			if(p[i][j]=='#') Put(i, j, 0);
		}
	}
	while (!q.empty()) {
            pair<int,int>  t= q.front();    q.pop();
		for (i = 0; i < 4; i++) Put(t.first + dx[i], t.second + dy[i], D[t.first][t.second] + 1);
	}
	printf("%d\n", ans);
}