The new cottage settlement is organized near the capital of Flatland. The construction company that is building the settlement has decided to paint some cottages pink and others — light blue. However, they cannot decide which cottages must be painted which color. The director of the company claims that the painting is nice if there is at least one pink cottage, at least one light blue cottage, and it is possible to draw a straight line in such a way that pink cottages are at one side of the line, and light blue cottages are at the other side of the line (and no cottage is on the line itself). The main architect objects that there are several possible nice paintings.
Help them to find out how many different nice paintings are there

//qscqesze

#include <cstdio>

#include <cmath>

#include <cstring>

#include <ctime>

#include <iostream>

#include <algorithm>

#include <set>

#include <vector>

#include <sstream>

#include <queue>

#include <typeinfo>

#include <fstream>

#include <map>

#include <stack>

typedef long long ll;

using namespace std;

//freopen("D.in","r",stdin);

//freopen("D.out","w",stdout);

#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)

#define maxn 200001

#define mod 10007

#define eps 1e-9

int Num;

char CH[];

//const int inf=0x7fffffff;   //нчоч╢С

const int inf=0x3f3f3f3f;

inline ll read()

{

    ll x=,f=;char ch=getchar();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

//**************************************************************************************

struct node

{

    double x,y;

};

int gcd(int x,int y)

{

    return y==?x:gcd(y,x%y);

}

map< pair<int,int> ,int> H;

node a[maxn];

int main()

{

    freopen("cottages.in","r",stdin);

    freopen("cottages.out","w",stdout);

    int n=read();

    for(int i=;i<n;i++)

        cin>>a[i].x>>a[i].y;

    int ans=;

    for(int i=;i<n;i++)

    {

        H.clear();

        for(int j=i+;j<n;j++)

        {

            int aa=a[i].x-a[j].x;

            int bb=a[i].y-a[j].y;

            int cc=gcd(aa,bb);

            if(H[make_pair(aa/cc,bb/cc)]==)

            {

                ans++;

                H[make_pair(aa/cc,bb/cc)]=;

            }

        }

    }

    cout<<ans*<<endl;

}