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

Codeforces Round #308 (Div. 2) D. Vanya and Triangles，
D. Vanya and Trianglestime limit per test4 secondsmemory limit per test512 megabytesinputstandard inputoutputstandard 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

40 01 12 02 2

output

3

input

30 01 12 0

output

1

input

11 1

output

0

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.

題目要求是求出n個點所能組成的三角形的個數，我們知道，若都能組成三角形，則總個數為c(n,3)個，現在目標就是求出不能組成三角形的個數，則

可以這樣求，以一個點為起點，其它點與這個點的斜率儲存在map中，則對於每個則有c(m,2)個不能組成三角形，因為重複算了三次，最後要除以3，

還要用long long 儲存結果，儲存斜率可以用分數a/b,這裡a,b互質，分數可以用a*N+b一個整數儲存起來，就沒有誤差了！

#define INF9000000000000000000#define EPS(double)1e-9#define mod1000000007#define PI3.14159265358979//*******************************************************************************/#endif#define N 2005#define MOD 1000000007int n;struct node{    int x,y;    node(int xx,int yy){        x = xx,y = yy;    }    node(){        x = 0;y=0;    }};node nodes[N];map<int,int> mymaps;map<int,int>::iterator it;int get(int a,int b){//a/b    if(a == 0) return 0;    if(b == 0) return N * N + N;    int sa = a>=0?1:-1,sb= b>=0?1:-1;    sa = sa * sb;sb = 1;    a = abs(a);b=abs(b);    int g = gcd(a,b);    a/=g;b/=g;    return sa * a * N + sa * b;}int main(){    //printf("%d",get(-1,2));    while (S(n) != EOF)    {        FI(n){            S2(nodes[i].x,nodes[i].y);        }        long long ans = (long long)n*((long long)n-1)*((long long)n-2)/6,anst = 0;        FI(n){            mymaps.clear();            FJ(n){                if(i != j){                    mymaps[get(nodes[j].y - nodes[i].y,nodes[j].x - nodes[i].x)]++;                }            }            for(it = mymaps.begin();it!=mymaps.end();it++){                int num = it->second;                anst += num>=2? num*(num-1)/2:0;            }        }        cout<<ans - anst/3<<endl;    }    return 0;}

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