D. Vanya and Triangles（Codeforces Round #308 (Div. 2)）

D. Vanya and Triangles time limit per test 4 seconds memory limit per test 512 megabytes input standard input output standard output

Vanya got bored and he painted n distinct points on the plane. After that he connected all the points pairwise and saw that as a result many triangles were formed with vertices in the painted points. He asks you to count the number of the formed triangles with the non-zero area.

Input

The first line contains integer n (1 ≤ n ≤ 2000) — the number of the points painted on the plane.

Next n lines contain two integers each xi, yi ( - 100 ≤ xi, yi ≤ 100) — the coordinates of the i-th point. It is guaranteed that no two given points coincide.

Output

In the first line print an integer — the number of triangles with the non-zero area among the painted points.

Sample test(s) input

4
0 0
1 1
2 0
2 2
      

output

3
      

input

3
0 0
1 1
2 0
      

output

1
      

input

1
1 1
      

output

Note

Note to the first sample test. There are 3 triangles formed: (0, 0) - (1, 1) - (2, 0); (0, 0) - (2, 2) - (2, 0); (1, 1) - (2, 2) - (2, 0).

Note to the second sample test. There is 1 triangle formed: (0, 0) - (1, 1) - (2, 0).

Note to the third sample test. A single point doesn't form a single triangle.

給你二維坐标下的n個點，問一共能構成多少個面積不為0的三角形。

轉載請注明出處：尋找&星空の孩子 

題目連結：http://codeforces.com/contest/552/problem/D

#include<cstdio>
#include<cmath>
#include<iostream>
#define PI acos(-1.0)
using namespace std;

struct Point
{
    double x,y;
    Point(double x=0,double y=0):x(x),y(y){}

};

typedef Point Vector;


Vector operator + (Vector A,Vector B){return Vector(A.x+B.x,A.y+B.y);}


Vector operator - (Point A,Point B){return Vector(A.x-B.x,A.y-B.y);}


Vector operator * (Vector A,double p){return Vector(A.x*p,A.y*p);}


Vector operator / (Vector A,double p){return Vector(A.x/p,A.y/p);}


bool operator < (const Point& a,const Point& b){return a.x<b.x||(a.x==b.x && a.y<b.y);}


const double eps = 1e-10;

int dcmp(double x){if(fabs(x)<eps)return 0;else return x < 0 ? -1 : 1;}

bool operator == (const Point& a,const Point& b){return dcmp(a.x-b.x)==0 && dcmp(a.y-b.y)==0;}


double Dot(Vector A,Vector B){return A.x*B.x+A.y*B.y;}
double length(Vector A){return sqrt(Dot(A,A));}
double Angle(Vector A,Vector B){return acos(Dot(A,B)/length(A)/length(B));}

double Cross(Vector A,Vector B){return A.x*B.y-B.x*A.y;}
double Area2(Point A,Point B,Point C){return Cross(B-A,C-A);}


inline Point read_point(Point &P)
{
    scanf("%lf%lf",&P.x,&P.y);
    return P;
}
int main()
{
    int n;
    Point po[2005];
    scanf("%d",&n);
    for(int i=0;i<n;i++)
        read_point(po[i]);
    if(n<3){printf("0\n");return 0;}
    int cnt=0;
    for(int i=0;i<n;i++)
    {
        for(int j=i+1;j<n;j++)
        {
            for(int k=j+1;k<n;k++)
            {
                if(Area2(po[i],po[j],po[k])!=0) cnt++;
            }
        }
    }
    printf("%d\n",cnt);
    return 0;
}