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>
#define pr pair<int,int>
using namespace std;

queue<pr>q;
int H,W;
char ss[1005][1005];
int vis[1005][1005];
int dx[4] = {0,-1,1,0};
int dy[4] = {1,0,0,-1};
int main() {
	scanf("%d%d",&H,&W);
	for(int i=1;i<=H;i++) {
			
		scanf("%s",&ss[i][1]);
			for(int j=1;j<=W;j++)
			if(ss[i][j]=='#') {
				vis[i][j] = 1;
				q.push(pr(i,j));
			}
		
	}
	int ans = 0;
	while(q.size()) {
		int x = q.front().first; int y = q.front().second;
		q.pop();
		ans = max(ans,vis[x][y]-1);
		for(int i=0;i<4;i++) {
			int nx = x+dx[i]; int ny = y+dy[i];
			if(nx>=1&&nx<=H&&ny>=1&&ny<=W&&!vis[nx][ny]) {
				vis[nx][ny] = vis[x][y]+1;
				q.push(pr(nx,ny));
			}
		}
	}
	cout<<ans;
}