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;
char s[1010][1010];
typedef pair<int,int> pii;
int d[1010][1010];

queue<pii>q;
int main() {
	scanf("%d%d",&N,&M);
	for (int i=1;i<=N;i++) {
		scanf("%s",s[i] + 1);
	}
	for (int i=1;i<=N;i++) {
		for (int j=1;j<=M;j++) {
			d[i][j] = -1;
			if (s[i][j] == '#') {
				d[i][j] = 0;
				q.push(pii(i,j));
			}
		}
	}
	int ans = 0;
	int dx[4] = {0,1,0,-1}, dy[4] = {1,0,-1,0};
	while(!q.empty()) {
		auto now = q.front();
		q.pop();
		int x = now.first, y = now.second;
		ans = max(ans,d[x][y]);
		for (int i=0;i<4;i++) {
			int nx = x + dx[i];
			int ny = y + dy[i];
			if (nx < 1 || ny < 1 || nx > N || ny > M || d[nx][ny] != -1) {
				continue;
			}
			d[nx][ny] = d[x][y] + 1;
			q.push(pii(nx,ny));
		}
	}
	printf("%d\n",ans);
}