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;
const int N=1010;
int n,m,cnt,t1,t2,vis[N][N];
struct data{int x,y;}A[N*N],B[N*N];
char s[N];
int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0};
int main(){
//	freopen("A.in","r",stdin);
//	freopen("A.out","w",stdout);
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;++i){
		scanf("%s",s+1);
		for(int j=1;j<=m;++j){
			vis[i][j]=s[j]=='#';
			if(!vis[i][j])continue;
			A[++t1]=(data){i,j};
		}
	}
	int idx=0;
	while(1){
		for(int i=1;i<=t1;++i){
			cnt++;
			int x=A[i].x,y=A[i].y;
			for(int j=0;j<4;++j){
				int nx=x+dx[j],ny=y+dy[j];
				if(nx<1||ny<1||nx>n||ny>m||vis[nx][ny])continue;
				vis[nx][ny]=1;
				B[++t2]=(data){nx,ny};
			}
		}
		++idx;if(cnt==n*m)break;
		for(int j=1;j<=t2;++j)A[j]=B[j];t1=t2;t2=0;
	}
	cout << idx - 1 << endl; 
	return 0;
}