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<cstdio>
#include<iostream>
#include<queue>
using namespace std;
const int maxn=1111;
int dis[maxn][maxn],A[maxn][maxn],X[5]={0,-1,1,0,0},Y[5]={0,0,0,-1,1};
char ss[maxn];
bool vis[maxn][maxn];
struct hhh
{
	int x,y;
	hhh(int a=0,int b=0)
	{
		x=a,y=b;
	}
};
queue<hhh> Q;
int main()
{
	int N,M;scanf("%d%d",&N,&M);
	for(int i=1;i<=N;++i)
	{
		scanf("%s",ss+1);
		for(int j=1;j<=M;++j) A[i][j]=(ss[j]=='#');
	}
	for(int i=1;i<=N;++i)for(int j=1;j<=M;++j)if(A[i][j]) Q.push(hhh(i,j)),vis[i][j]=true;
	while(!Q.empty())
	{
		hhh q=Q.front();Q.pop();
		for(int i=1;i<=4;++i)
		{
			int xx=q.x+X[i],yy=q.y+Y[i];
			if(!xx||!yy||xx>N||yy>M||vis[xx][yy]) continue;
			dis[xx][yy]=dis[q.x][q.y]+1;
			Q.push(hhh(xx,yy)),vis[xx][yy]=true;
		}
	}
	int ans=0;
	for(int i=1;i<=N;++i)for(int j=1;j<=M;++j) ans=max(ans,dis[i][j]);
	printf("%d",ans);
	return 0;
}