Assume the coasting is an infinite straight line. Land is in one side of coasting, sea in the other. Each small island is a point locating in the sea side. And any radar installation, locating on the coasting, can only cover d distance, so an island in the sea can be covered by a radius installation, if the distance between them is at most d.

We use Cartesian coordinate system, defining the coasting is the x-axis. The sea side is above x-axis, and the land side below. Given the position of each island in the sea, and given the distance of the coverage of the radar installation, your task is to write a program to find the minimal number of radar installations to cover all the islands. Note that the position of an island is represented by its x-y coordinates.

Figure A Sample Input of Radar Installations

Input

The input consists of several test cases. The first line of each case contains two integers n (1<=n<=1000) and d, where n is the number of islands in the sea and d is the distance of coverage of the radar installation. This is followed by n lines each containing two integers representing the coordinate of the position of each island. Then a blank line follows to separate the cases.

The input is terminated by a line containing pair of zeros

Output

For each test case output one line consisting of the test case number followed by the minimal number of radar installations needed. "-1" installation means no solution for that case.

Sample Input

3 2

1 2

-3 1

2 1

1 2

0 2

0 0

Sample Output

Case 1: 2

Case 2: 1

贪心：注意重点

#include<iostream>

#include<cstring>

#include<algorithm>

#include<cstdio>

#include<cmath>

using namespace std;

int f[1010];

struct aa

{

    double x,y;

}a[1010];

int cmp(aa q,aa w)

{

    return q.x<w.x;

}

int main()

{

    int k,n,m,cnt,ka=0;

    while(cin>>n>>m)

    {   if(n==0&&m==0)break;

        ka++;

        int flat=1;

        k=0;

        double x,y,c,mm=(double)m*m;

        for(int i=1;i<=n;i++)

        {

          cin>>x>>y;

          if(m>=y)

          {c=sqrt(mm-y*y);

           a[++k].x=x-c;

           a[k].y=x+c;

          }

          else

         {

             flat=0;

          }

        }

        getchar();

        if(flat)

        {sort(a+1,a+k+1,cmp);

        double now = a[1].y;

         cnt = 1;

        for(int i=2;i<=k;i++)

        {

          if(a[i].y<now)

            now=a[i].y;

          else if(a[i].x>now)

          {

              cnt++;

              now=a[i].y;

          }

        }

        }

        else

        cnt=-1;

        cout<<"Case "<<ka<<": "<<cnt<<endl;

    }

    return 0;

}