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>
#define Tp template<typename Ty>
#define Ts template<typename Ty,typename... Ar>
#define Reg register
#define RI Reg int
#define Con const
#define CI Con int&
#define I inline
#define W while
#define N 1000
using namespace std;
const int dx[4]={0,0,1,-1},dy[4]={1,-1,0,0};
int n,m,vis[N+5][N+5];char s[N+5][N+5];struct Data {int x,y,t;}q[N*N+5];
int main()
{
	RI i,j,H=1,T=0;for(scanf("%d%d",&n,&m),i=1;i<=n;++i)
		for(scanf("%s",s[i]+1),j=1;j<=m;++j) s[i][j]=='#'&&(q[++T]=(Data){i,j,0},vis[i][j]=1);
	Data k;RI nx,ny;W(H<=T) for(k=q[H++],i=0;i^4;++i) (nx=k.x+dx[i])&&nx<=n&&
		(ny=k.y+dy[i])&&ny<=m&&!vis[nx][ny]&&(q[++T]=(Data){nx,ny,k.t+1},vis[nx][ny]=1);
	return printf("%d\n",q[T].t),0;
}