Catching Fish

Ignatius likes catching fish very much. He has a fishnet whose shape is a circle of radius one. Now he is about to use his fishnet to catch fish. All the fish are in the lake, and we assume all the fish will not move when Ignatius catching them. Now Ignatius wants to know how many fish he can catch by using his fishnet once. We assume that the fish can be regard as a point. So now the problem is how many points can be enclosed by a circle of radius one. 

Note: If a fish is just on the border of the fishnet, it is also caught by Ignatius. 

 

Input

The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow. 
Each test case starts with a positive integer N(1<=N<=300) which indicate the number of fish in the lake. Then N lines follow. Each line contains two floating-point number X and Y (0.0<=X,Y<=10.0). You may assume no two fish will at the same point, and no two fish are closer than 0.0001, no two fish in a test case are approximately at a distance of 2.0. In other words, if the distance between the fish and the centre of the fishnet is smaller 1.0001, we say the fish is also caught. 

 

Output

For each test case, you should output the maximum number of fish Ignatius can catch by using his fishnet once. 

 

Sample Input

4 3 6.47634 7.69628 5.16828 4.79915 6.69533 6.20378 6 7.15296 4.08328 6.50827 2.69466 5.91219 3.86661 5.29853 4.16097 6.10838 3.46039 6.34060 2.41599 8 7.90650 4.01746 4.10998 4.18354 4.67289 4.01887 6.33885 4.28388 4.98106 3.82728 5.12379 5.16473 7.84664 4.67693 4.02776 3.87990 20 6.65128 5.47490 6.42743 6.26189 6.35864 4.61611 6.59020 4.54228 4.43967 5.70059 4.38226 5.70536 5.50755 6.18163 7.41971 6.13668 6.71936 3.04496 5.61832 4.23857 5.99424 4.29328 5.60961 4.32998 6.82242 5.79683 5.44693 3.82724 6.70906 3.65736 7.89087 5.68000 6.23300 4.59530 5.92401 4.92329 6.24168 3.81389 6.22671 3.62210

 

Sample Output
2 5 5 11

枚举两个点，由这两个在圆上的点确定圆心的坐标，然后看在这个园中有多少个点

#include<iostream>
#include <cstdio>
#include<cstring>
#include<algorithm>
#include <cmath>
using namespace std;
 struct point
 {
     double x,y;
 };
 
 double dis(point a,point b)
   {
      return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
   }
point center(point a,point b)//求出圆心的位置
  {
      point s,e,m;
     s.x=b.x-a.x;
     s.y=b.y-a.y;
     m.x=(a.x+b.x)/2.0;///求a，b的中点
     m.y=(a.y+b.y)/2.0;
     double d=dis(a,m);
     double an=sqrt(1.0-d);
    if(fabs(s.y)<1e-8)///如果ab与x轴平行
    {
        e.x=m.x;
        e.y=m.y+an;
    }
   else
    {
        double ang=atan(-s.x/s.y);//求该线段垂直平分线与x轴所形成的角
        e.x=m.x+an*cos(ang);
        e.y=m.y+an*sin(ang);
    }
   return e;
  }
 
 
int main()
{
   int T,n,i,j,k;
    point p[310],c;
   scanf("%d",&T);
  while(T--)
  {
    scanf("%d",&n);
   for(i=0;i<n;i++)
     scanf("%lf%lf",&p[i].x,&p[i].y);
 
   int ans=1;
   for(i=0;i<n-1;i++)
     for(j=i+1;j<n;j++)
   {
       if(dis(p[i],p[j])<=4.0)//枚举两个点
         {
              c=center(p[i],p[j]);
 
            int tmp=0;
            for(k=0;k<n;k++)
            {
                if(sqrt(dis(c,p[k]))<=1.000001)
                    tmp++;
            }
            if(ans<tmp) ans=tmp;
         }
 
   }
   printf("%d\n",ans);
  }
    return 0;
}