Problem: 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

Solution (in CPP):

Solution:
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1005,M=1e6+5;
struct Node{
  int x,y;
  Node(int x=0,int y=0):x(x),y(y) {}
}q[M];
int n,m,he,tl,dis[N][N];
bool vis[N][N]; char s[N];
int main()
{
  scanf("%d%d",&n,&m);
  for(int i=1;i<=n;i++)
    {
      scanf("%s",s+1);
      for(int j=1;j<=m;j++)
	if(s[j]=='#')
	  { vis[i][j]=1;q[++tl]=Node(i,j);}
    }
  while(he<tl)
    {
      Node cr=q[++he]; int x=cr.x,y=cr.y,k=dis[x][y];
      if(x>1&&!vis[x-1][y])
	vis[x-1][y]=1,q[++tl]=Node(x-1,y),dis[x-1][y]=k+1;
      if(y>1&&!vis[x][y-1])
	vis[x][y-1]=1,q[++tl]=Node(x,y-1),dis[x][y-1]=k+1;
      if(x<n&&!vis[x+1][y])
	vis[x+1][y]=1,q[++tl]=Node(x+1,y),dis[x+1][y]=k+1;
      if(y<m&&!vis[x][y+1])
	vis[x][y+1]=1,q[++tl]=Node(x,y+1),dis[x][y+1]=k+1;
    }
  printf("%d\n",dis[q[tl].x][q[tl].y]);
  return 0;
}