Eureka

Professor Zhang draws n points on the plane, which are conveniently labeled by 1,2,...,n. The i-th point is at (xi,yi). Professor Zhang wants to know the number of best sets. As the value could be very large, print it modulo 109+7.

A set P (P contains the label of the points) is called best set if and only if there are at least one best pair in P. Two numbers u and v (u,v∈P,u≠v) are called best pair, if for every w∈P, f(u,v)≥g(u,v,w), where f(u,v)=(xu−xv)2+(yu−yv)2−−−−−−−−−−−−−−−−−−√ and g(u,v,w)=f(u,v)+f(v,w)+f(w,u)2.

 

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first line contains an integer n (1≤n≤1000) -- then number of points.

Each of the following n lines contains two integers xi and yi (−109≤xi,yi≤109) -- coordinates of the i-th point.

 

Output

For each test case, output an integer denoting the answer.

 

Sample Input

3 3 1 1 1 1 1 1 3 0 0 0 1 1 0 1 0 0

 

Sample Output

4 3 0

 

 

#include <stdio.h>
#include <string.h>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
using namespace std;
typedef long long LL;
#define MAXM 1000000004
#define pi 1000000007

map<pair<int ,int> ,int>x;
map<pair<int ,int>  ,int >::iterator it;




struct aa
{
    long long x,y;

} a[1001];

int cmp(aa t1,aa t2)
{
    return t1.x==t2.x?t1.y<t2.y:t1.x<t2.x;
}
LL gcd(LL a,LL b)
{
    return b==0?a:gcd(b,a%b);
}

LL qe(LL a, LL b)
{
    LL ans = 1;
    a = a%pi;
    while (b > 0)
    {
        if (b % 2 == 1)
            ans = (ans*a) % pi;
        b = b / 2;
        a = (a*a) % pi;
    }
    return ans;
}

int main()
{
  //  init();
    int t,i,j,k;
    LL ans=0;
    cin>>t;
    pair <int ,int>h;
    while(t--)
    {
        cin>>k;
        ans=0;
        for(i=1; i<=k; i++)
            scanf("%lld %lld",&a[i].x,&a[i].y);
        sort(a+1,a+1+k,cmp);
        for(i=1; i<k; i++)
        {

            LL sum=1;
            x.clear();
            for(j=i+1; j<=k; j++)
            {
                LL x1,y1;
                x1=a[j].x-a[i].x;
                y1=a[j].y-a[i].y;
                if(x1==0&&y1==0)
                {
                    sum++;
                    continue;
                }
                else
                {
                    LL se;
                    se=gcd(x1,y1);
                    x1/=se;
                    y1/=se;
                    h=make_pair(x1,y1);
                }
                x[h]++;
            }
            if(sum>1)
            {
                ans=(ans+qe(2,sum)-1-sum)%pi;
            }

            LL u=0;
            u=qe(2,sum)-1;
            // for(it=vis.begin(); it!=vis.end(); it++)
            for(it=x.begin(); it!=x.end(); it++)
            {
                LL v,w=0;
                v=it->second;
                v=qe(2,v);
                ans=(ans+(u)*(v-1))%pi;
            }

while(i!=k&&a[i].x==a[i+1].x&&a[i].y==a[i+1].y)
{
    i++;
}
        }
        printf("%I64d\n",ans);

    }
}