#include<iostream>
#include<string.h>
#include<string>
#include<algorithm>
#include<math.h>
#include<string>
#include<string.h>
#include<vector>
#include<utility>
#include<map>
#include<queue>
#include<set>
#define mx 0x3f3f3f3f
#define ll long long
using namespace std;
const int N = 100010;
int flag;
double ans1,ans2,yy;
struct Point//定义点的结构体
{
    double x, y;
};
struct stline//定义边的结构体
{
    Point a, b;
} line[105];

bool cmp(Point a, Point b)
{
    return a.y < b.y;
}
int dblcmp(double a, double b)
{
    if (fabs(a - b) <= 1E-6) return 0;
    if (a > b) return 1;
    else return -1;
}
//***************点积判点是否在线段上***************
double dot(double x1, double y1, double x2, double y2) //点积
{
    return x1 * x2 + y1 * y2;
}

int point_on_line(Point a, Point b, Point c) //求a点是不是在线段bc上，>0不在，=0与端点重合，<0在。
{
    return dblcmp(dot(b.x - a.x, b.y - a.y, c.x - a.x, c.y - a.y), 0);
}
//**************************************************
double cross(double x1, double y1, double x2, double y2)
{
    return x1 * y2 - x2 * y1;
}
double ab_cross_ac(Point a, Point b, Point c) //ab与ac的叉积
{
    return cross(b.x - a.x, b.y - a.y, c.x - a.x, c.y - a.y);
}
int ab_cross_cd (Point a,Point b,Point c,Point d) //求ab是否与cd相交，交点为p。1规范相交，0交点是一线段的端点，-1不相交。
{
    double s1,s2,s3,s4;
    int d1,d2,d3,d4;
    Point p;
    d1=dblcmp(s1=ab_cross_ac(a,b,c),0);
    d2=dblcmp(s2=ab_cross_ac(a,b,d),0);
    d3=dblcmp(s3=ab_cross_ac(c,d,a),0);
    d4=dblcmp(s4=ab_cross_ac(c,d,b),0);

//如果规范相交则求交点
    if ((d1^d2)==-2 && (d3^d4)==-2)
    {
        p.x=(c.x*s2-d.x*s1)/(s2-s1);
        p.y=(c.y*s2-d.y*s1)/(s2-s1);
        return 1;
    }

//如果不规范相交
    if (d1==0 && point_on_line(c,a,b)<=0)
    {
        p=c;
        return 0;
    }
    if (d2==0 && point_on_line(d,a,b)<=0)
    {
        p=d;
        return 0;
    }
    if (d3==0 && point_on_line(a,c,d)<=0)
    {
        p=a;
        return 0;
    }
    if (d4==0 && point_on_line(b,c,d)<=0)
    {
        p=b;
        return 0;
    }
//如果不相交
    return -1;
}
int main()
{
    int t;
    while(scanf("%d", &t)&&t)
    {
        int cnt=0;
        for(int i=1;i<=t;i++)
            scanf("%lf%lf%lf%lf", &line[i].a, &line[i].a.y, &line[i].b.x, &line[i].b.y);
        for(int i=1;i<=t;i++)
        {
            for(int j=i+1;j<=t;j++)
            {
                if(ab_cross_cd(line[i].a, line[i].b, line[j].a, line[j].b)!=-1)
                    cnt++;
            }
        }
        printf("%d\n",cnt);
    }
    return 0;
}