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

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

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.

Case 1: 2

Case 2: 1

    先算出每个岛放探测雷达的x坐标范围，b为雷达探测半径，岛的坐标为（x，y），雷达的坐标范围是（x-sqrt(d*d-y*y),x+sqrt(d*d-y*y)），定义sum=1，i从1开始，每个坐标范围左右边界与上一个坐标范围的右边界比较，具体看代码。

 #include<cstdio>

 #include<algorithm>

 #include<cmath>

 using namespace std;

 struct stu

 {

     double l,r;

 } st[];

 bool cmp(stu a,stu b)

 {

     return a.l<b.l;

 }

 int main()

 {

     int a,s,k=,sum,i;

     double b,x,y,sky;

     while(scanf("%d %lf",&a,&b) &&!(a==&&b==))

     {

         s=;

         k++;

         for(i =  ; i < a ; i++)

         {

             scanf("%lf %lf",&x,&y);

             if(b <  || y > b)

             {

                 s=-;

             }

             st[i].l=x-sqrt(b*b-y*y);

             st[i].r=x+sqrt(b*b-y*y);

         }

         if(s == -) printf("Case %d: -1\n",k);

         else

         {

             sort(st,st+a,cmp);

             sum=;

             sky=st[].r;

             for(i =  ; i < a ; i++)

             {

                 if(st[i].r <= sky)

                 {

                     sky=st[i].r;

                 }

                 else

                 {

                     if(st[i].l > sky)

                     {

                         sky=st[i].r;

                         sum++;

                     }

                 }

             }

             printf("Case %d: %d\n",k,sum);

         }

     }

 }