C - Line-line Intersection Gym - 102220C（线段相交）

There are n lines l1,l2,…,ln

on the 2D-plane.

Staring at these lines, Calabash is wondering how many pairs of (i,j)

that 1≤i<j≤n and li,lj

share at least one common point. Note that two overlapping lines also share common points.

Please write a program to solve Calabash's problem.

Input

The first line of the input contains an integer T(1≤T≤1000)

, denoting the number of test cases.

In each test case, there is one integer n(1≤n≤100000)

in the first line, denoting the number of lines.

For the next n

lines, each line contains four integers xai,yai,xbi,ybi(|xai|,|yai|,|xbi|,|ybi|≤109). It means li passes both (xai,yai) and (xbi,ybi). (xai,yai) will never be coincided with (xbi,ybi)

.

It is guaranteed that ∑n≤106

.

Output

For each test case, print a single line containing an integer, denoting the answer.

Example

Input

3
2
0 0 1 1
0 1 1 0
2
0 0 0 1
1 0 1 1
2
0 0 1 1
0 0 1 1

Output

1
0
1

题解：两条直线不平行必相交，若平行：若重合答案加1，否则不算。

我们用两个map来刻画直线的特性，mp1刻画a*x+b*y的直线系有多少个，mp2刻画a*x+b*y=c这一条直线有多少个。

假设当前直线与之前的线段都相交，那么我们需要减去与这条直线平行而不重合的直线。即ans+=i-1+mp2[]-mp1[].

#include<iostream>
#include<cstring>
#include<string>
#include<queue>
#include<stack>
#include<algorithm>
#include<stdio.h>
#include<map>
#include<set>
using namespace std;
typedef long long ll;
typedef pair<ll,ll>P;
typedef pair<pair<ll,ll>,ll>Pi;
const int maxn=100010;
map<pair<ll,ll>,ll>mp1;//两个参数a,b，代表形如a*x+b*y=c（c任意）的直线有多少个
map<pair<pair<ll,ll>,ll>,ll>mp2;//三个参数a,b,c,代表形如a*x+b*y=c的直线有多少个，即相同直线有多少个
ll cnt,ans,n;
int main()
{
    ios::sync_with_stdio(0);
    int T;
    cin>>T;
    while(T--){
        ans=cnt=0;
        cin>>n;
        mp1.clear(),mp2.clear();
        for(int i=1;i<=n;i++){
            ll x1,x2,y1,y2;
            cin>>x1>>y1>>x2>>y2;
            ll a=x1-x2,b=y1-y2,c=x1*y2-x2*y1;
            ll g=__gcd(a,b);
            a/=g;b/=g;c/=g;
            mp1[P(a,b)]++;
            mp2[Pi(P(a,b),c)]++;
            ans+=i-1+mp2[Pi(P(a,b),c)]-mp1[P(a,b)];//假设与前i-1条边都相交，需要减去与他平行而不重合的线段
        }
        cout<<ans<<endl;
    }
    return 0;
}