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<cstring>
const int N=1005,dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
int n,m,he,ta,q[N*N][2],dis[N][N];char s[N][N];bool vis[N][N];
int get(int x,int y){return x*N+y;}
int main(){
	memset(dis,63,sizeof(dis));
	memset(vis,1,sizeof(vis));
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;++i)scanf("%s",s[i]+1);
	for(int i=1;i<=n;++i)for(int j=1;j<=m;++j)if(s[i][j]=='#')q[++ta][0]=i,q[ta][1]=j,dis[i][j]=0;else vis[i][j]=0;
	while(he!=ta){
		int x=q[++he][0],y=q[he][1];
		for(int i=0;i<4;++i){
			int nx=x+dir[i][0],ny=y+dir[i][1];
			if(!vis[nx][ny]){
				vis[nx][ny]=1;
				dis[nx][ny]=dis[x][y]+1;
				q[++ta][0]=nx,q[ta][1]=ny;
			}
		}
	}
	printf("%d\n",dis[q[ta][0]][q[ta][1]]);
	return 0;
}