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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 21 2-3 12 11 20 20 0

Sample Output

Case 1: 2Case 2: 1
题目解析：
      输入的是点n个的数目，和雷达m的范围，后面再加上n行点的位置，要求雷达在x轴（x轴指的是海平面，以下同理）上面。题目求的是，要想将所有的点覆盖的雷达数目，求最小的雷达数目。
主要的思路就是将这个题目转换一下去求。题目上面说，雷达只会在x轴上面，那么我们就可以去算一下大致的几种情况：
           第一种，雷达全面覆盖的到。那么转换一下思路。
           第二种，存在几个特殊的，比较高的位置，雷达无法覆盖，就是雷达最高为m，但是他的位置高度超过了m，所以直接输出-1.
思路就是我们可以把每一个点转换一下，看成一个圆，去思考一下。如果点可以被覆盖的到，那么圆与x轴就会相交或者相切，以半径为雷达m范围的圆。如果他们不相交，那么就是说雷达扫描不到，因为雷达只在x轴上面。
如图：
 
在这里我们看见了因为相交的部分不重合，所以要想在x轴上面安雷达覆盖这两个点，我们至少需要两个。
如图的思想，我们再加一个点所构成的圆。在此注明一点就是，在x轴上面的范围就是表明雷达如果要覆盖这点，那么他就必须在这个范围里面。因为二点，一必须在x轴上面，二他是以雷达范围的圆，不在就说明点与雷达的范围超过了雷达扫描的范围。
在此图，我们可以更好的看出如果，范围重合了，就说明他们可以用一个雷达去扫描得到。
最后得到的其实就是这个图了，求的就是公共覆盖的就用一点，不是公共覆盖的就开辟，把一个范围不断的缩小，缩小的区域就是共同的雷达，不在缩小的区域类，就代表着，存在一点和你不含有公共区域。
所以题目的思路就清晰了，先把点转化为在x轴上面的范围，求范围的话，根据数学的公式就可以求出来了。[x-根号下(r*r-y*y)，x+根号下(r*r-y*y)]，这个范围就可以了。最后题目就转换成了求公共覆盖范围用一个雷达，不在就新建一个雷达。
代码如下：
#include<iostream>#include<cstdio>#include<algorithm>#include<cmath>using namespace std;struct node{    double left;    double right;} a[1003];bool cmp(node n,node m){    return n.left<m.left;}int main(){    int n;double m;    int count=1;    while(scanf("%d%lf",&n,&m)!=EOF)    {        if(n==0 && m==0) break;        double k,t;        bool flag = false;        for(int i=0; i<n; i++)        {            scanf("%lf%lf",&k,&t);            a[i].left = k - sqrt((double)m*m-(double)t*t);            a[i].right = k + sqrt((double)m*m-(double)t*t);            if(t>m || m<0)                flag=true;        }        //超出范围，无法扫描到。        if(flag)        {            printf("Case %d: %d\n",count++,-1);        }        else        {            sort(a,a+n,cmp);double t=a[0].right;            int c=1;            for(int i=1; i<n; i++)            {                if(a[i].right<=t)                    t = a[i].right;                else                {                    if(a[i].left>t)                    {                        t=a[i].right;                        c++;                    }                }            }            printf("Case %d: %d\n",count++,c);        }    }    return 0;}

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