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 rep(i,n) for(int i=0;i<(n);i++)

using namespace std;

const int dx[]={1,0,-1,0},dy[]={0,-1,0,1};

int main(){
	int h,w; scanf("%d%d",&h,&w);
	static char B[1000][1001];
	rep(i,h) scanf("%s",B[i]);

	static int d[1000][1000];
	memset(d,-1,sizeof d);
	queue<pair<int,int>> Q;
	rep(i,h) rep(j,w) if(B[i][j]=='#') {
		d[i][j]=0;
		Q.emplace(i,j);
	}
	while(!Q.empty()){
		int y,x;
		tie(y,x)=Q.front(); Q.pop();

		rep(k,4){
			int yy=y+dy[k],xx=x+dx[k];
			if(0<=yy && yy<h && 0<=xx && xx<w && d[yy][xx]==-1){
				d[yy][xx]=d[y][x]+1;
				Q.emplace(yy,xx);
			}
		}
	}

	int ans=0;
	rep(i,h) rep(j,w) ans=max(ans,d[i][j]);
	printf("%d\n",ans);

	return 0;
}