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<bits/stdc++.h>
using namespace std;
#define mp make_pair
#define x first
#define y second
#define pii pair<int ,int>
const int N=1005;
queue<pii> Q;
int n,m,ans,flag[N][N],dx[4]={1,-1,0,0},dy[4]={0,0,1,-1};
char s[N][N];
int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++)scanf("%s",s[i]+1);
	memset(flag,-1,sizeof flag);
	for (int i=1;i<=n;i++)
		for (int j=1;j<=m;j++)
			if (s[i][j]=='#')Q.push(mp(i,j)),flag[i][j]=0;
	while (!Q.empty()){
		pii now=Q.front();Q.pop();
		for (int i=0;i<4;i++){
			pair<int ,int > New=mp(now.x+dx[i],now.y+dy[i]);
			if (New.x>0&&New.y>0&&New.x<=n&&New.y<=m&&flag[New.x][New.y]==-1){
				Q.push(New);
				flag[New.x][New.y]=flag[now.x][now.y]+1;
				ans=max(ans,flag[New.x][New.y]);
			}
		}
	}
	printf("%d\n",ans);
}