struct pt {
    T X, Y; // T 可以為 int 或是 double
}

#define pt pair<int, int>
#define X first
#define Y second

pt v1;
cin >> v1.X >> v1.Y;

struct pt{
    int x,y;
};

pt operator+(const pt &a, const pt &b) {
    return {a.x + b.x, a.y + b.y};
} //向量相加

pt operator-(const pt &a, const pt &b) {
    return {a.x - b.x, a.y - b.y};
} //向量相減

pt operator*(int k, const pt &a) {
    return {a.x * k, a.y * k};
} //向量乘法（係數積）
pt operator*(const pt &a, int k) {
    return {a.x * k, a.y * k};
} // 乘法有交換率

pt operator/(const pt &a, int k) {
    return {a.x / k, a.y / k};
} //向量除法 (係數除法)

int operator^(const pt &a, const pt &b) {
    return a.x * b.y - a.y * b.x;
} //向量外積cross

int operator*(const pt &a, const pt &b) {
    return a.x * b.x + a.y * b.y;
} //向量內積dot

inline bool operator == (const pt &a, const pt &b){
    if(abs(a.x-b.x) <= eps && abs(a.y - b.y) <= eps) return true;
    return false;
}

//方向函數ori，回傳OA向量與OB向量的方向
int ori( const Pt& a , const Pt& b, const Pt& o){	//方向函數
    double cross = ( a - o ) ^ ( b - o );
    if( fabs( cross ) < eps ) return 0;
    return cross > 0 ? 1 : -1;
}

bool onseg(pt a, pt b, pt o){         //o是否在ab線段上
    int cross = (a - o) ^ (b - o);    //是否平行
    int dot = (a - o) * (b - o);      //是否在線段中
    return (cross == 0)&&(dot <= 0);
}

bool banana(pt a, pt b, pt c, pt d){      //線段ab是否與cd相交
    if(onseg(a,b,c)||onseg(a,b,d)) return true;  //點c、d是否洽在線段ab上
    if(onseg(c,d,a)||onseg(c,d,b)) return true;  //點a、b是否洽在線段cd上
    return (ori(a,b,c)*ori(a,b,d) < 0 && ori(c,d,a)*ori(c,d,b) < 0);
}

pt banana_point(pt a, pt b, pt c, pt d){
    assert(banana(a, b, c, d));		//是否相交
    return a + ((a - c)^(d - c) * (b - a))/((d - c)^(b - a));
}

#include <bits/stdc++.h>
#define int long long
#define double long double
#define pii pair<int, int>
#define N 1005
#define INF 1000000005
#define x first
#define y second
#define IOS ios::sync_with_stdio(0),cin.tie(0)
using namespace std;
int n,m;
 
struct point{
    int x,y;
    bool operator == (point b){
        if(x == b.x && y == b.y)return true;
        return false;
    }
    point operator - (point b){return {x - b.x , y - b.y};}
    point operator + (point b){return {x + b.x , y + b.y};}
    int operator ^ (point b){return (x * b.y - y * b.x);}
    int operator * (point b){return (x * b.x + y * b.y);}
}pt[N];
 
bool onseg(point a,point b,point o){
    int cross = (a - o) ^ (b - o);
    int dot = (a - o) * (b - o);
    return (cross == 0) && (dot <= 0);
}
 
int dir(point a,point b,point o){
    int cross = (a - o) ^ (b - o);
    if(cross == 0)return 0;
    else if(cross > 0)return 1;
    else return -1;
}
 
bool inter(point a,point b,point c,point d){
    if(onseg(a,b,c) || onseg(a,b,d))return true;
    if(onseg(c,d,a) || onseg(c,d,b))return true;
    if(dir(a,b,c) * dir(a,b,d) < 0 && dir(c,d,a) * dir(c,d,b) < 0)
        return true;
    return false;
}
 
int solve(point cur){
    int sum = 0;
    for(int i=0;i<n;i++){
        if(onseg(pt[i],pt[i+1],cur) == 1)return -1;
        if(inter(pt[i],pt[i+1],cur,point{INF,cur.y}))sum++;
        point temp = pt[i].y > pt[i+1].y ? pt[i] : pt[i+1];
        if(temp.y == cur.y && temp.x > cur.x)sum--;
    }
    return sum;
}
 
signed main(){
    IOS;
    cin>>n>>m;
    for(int i=0;i<n;i++)cin>>pt[i].x>>pt[i].y;
    pt[n] = pt[0];
    for(int i=0;i<m;i++){
        point temp;cin>>temp.x>>temp.y;
        int ans = solve(temp);
        if(ans == -1)cout<<"BOUNDARY"<<"\n";
        else if(ans & 1)cout<<"INSIDE"<<"\n";
        else cout<<"OUTSIDE"<<"\n";
    }
}
 

bool cmp(pt a, pt b){
    return atan2(a.y,a.x) < atan2(b.y,b.x);
}
sort(p.begin(),p.end(),cmp);

bool cmp(pt a, pt b){
    bool f1 = a < pt{0,0};
    bool f2 = b < pt{0,0};
    if(f1 != f2)return f1 < f2;
    return (a ^ b) > 0;
    //逆時針將點進行極角排序，從270度開始逆時針
}
sort(p.begin(),p.end(),cmp);
//以id為原點進行極角排序

#include <bits/stdc++.h>
#define Orz ios::sync_with_stdio(0),cin.tie(0)
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define pii pair<int,int>
#define pdd pair<double,double>
#define int long long
#define ll long long
#define ld long double
#define N 100001
#define all(x) x.begin(),x.end()
#define eps 1e-9
#define x first
#define y second

using namespace std;

struct pt{
    int x,y;
    bool operator < (pt b){
        if(x == b.x)return y < b.y;
        return x < b.x;
    }
    bool operator > (pt b){
        if(x == b.x)return y > b.y;
        return x > b.x;
    }
    bool operator == (pt b){
        if(x-b.x == 0 && y-b.y == 0)return true;
        return false;
    }
    pt operator+(pt b) {return {x + b.x, y + b.y};} //向量相加
    pt operator-(pt b) {return {x - b.x, y - b.y};} //向量相減
    int operator^(pt b) {return x * b.y - y * b.x;} //向量外積cross
    int operator*(pt b) {return x * b.x + y * b.y;} //向量內積dot
};

vector<pt> p,temp,pp;
vector<int>  cnt;
int n,ans = 0;

bool cmp(pt a, pt b){
    bool f1 = a < pt{0,0};
    bool f2 = b < pt{0,0};
    if(f1 != f2)return f1 < f2;
    return (a ^ b) > 0;
    //逆時針將點進行極角排序，從270度開始逆時針
}

//O(n)枚舉每個點當直角情況
int solve(pt id){
    pp.clear();cnt.clear();temp.clear();
    for(pt i : p){
        pt cur = i - id;
        if(cur == pt{0,0})continue;
        temp.push_back(cur);
    }
    sort(all(temp),cmp);            //以id為原點進行極角排序
    pp.push_back(temp[0]);          //pp每一角度只存至多一個點
    cnt.push_back(1);               //考慮每個點共線情況
    int len = temp.size();
    rep(i,1,len-1){
        int cross = temp[i]^temp[i-1],dot = temp[i]*temp[i-1];
        if(cross == 0 && dot >= 0)cnt[cnt.size()-1] += 1;   //共線數量+=1
        else {pp.push_back(temp[i]);cnt.push_back(1);}      //非共線設定數量為1
    }
    len = pp.size();            //考慮橫跨一周的情況
    rep(i,0,len-1){             //雙指針i,p1可能會超過一圈
        pp.push_back(pp[i]);    //將點再繞一圈
        cnt.push_back(cnt[i]);
    }
    int ans = 0,p1 = 0;
    rep(i, 0, len-1){
        while(p1 < i+len && (pp[i]^pp[p1]) >= 0 && (pp[i]*pp[p1]) > 0)p1 += 1;
        //夾銳角的情況要p1+=1
        if((pp[i]^pp[p1]) > 0 && (pp[i]*pp[p1]) == 0)ans += cnt[i]*cnt[p1];
        //正向的直角三角形，若共線則兩者數量相乘
    }
    return ans;
}

signed main(){
    Orz;
    while(cin>>n){
        if(n == 0)break;
        p.assign(n,{0,0});
        rep(i,0,n-1)cin>>p[i].x>>p[i].y;
        
        int ans = 0;
        rep(i,0,n-1){
            ans += solve(p[i]);
        }
        cout<<ans<<endl;
    }
}

#include <bits/stdc++.h>
#define Orz ios::sync_with_stdio(0),cin.tie(0)
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define pii pair<int,int>
#define pdd pair<double,double>
#define int long long
#define ll long long
#define ld long double
#define N 100001
#define all(x) x.begin(),x.end()
#define eps 1e-9
#define x first
#define y second

using namespace std;

struct pt{
    int x,y;
    bool operator < (pt b){
        if(x == b.x)return y < b.y;
        return x < b.x;
    }
    bool operator > (pt b){
        if(x == b.x)return y > b.y;
        return x > b.x;
    }
    bool operator == (pt b){
        if(x-b.x == 0 && y-b.y == 0)return true;
        return false;
    }
    pt operator+(pt b) {return {x + b.x, y + b.y};} //向量相加
    pt operator-(pt b) {return {x - b.x, y - b.y};} //向量相減
    int operator^(pt b) {return x * b.y - y * b.x;} //向量外積cross
    int operator*(pt b) {return x * b.x + y * b.y;} //向量內積dot
};

vector<pt> p;
vector<int> cnt;
int n,ans = 0;

bool cmp(pt a, pt b){
    bool f1 = a < pt{0,0};
    bool f2 = b < pt{0,0};
    if(f1 != f2)return f1 < f2;
    return (a ^ b) > 0;
    //逆時針將點進行極角排序，從270度開始逆時針
}

//用cnt[i]統計區間長度為i的線段數量
void solve(pt id){
    vector<pt> pp;
    for(auto i : p){                         //以id為原點
        pt cur = i-id;
        if(cur == pt{0,0})continue;
        if(cur.y < 0){cur.x = -cur.x;cur.y = -cur.y;}
        if(cur.x < 0 && cur.y==0){cur.x = -cur.x;}
        pp.push_back(cur);
    }
    sort(all(pp),cmp);                      //將id當作原點進行排序
    int p1 = 0,p2 = 0,len = pp.size();      //雙指針找共線區間
    while(p1 < n-1){                        //最大化區間
        while(p2+1 < len && (pp[p1]^pp[p2+1]) == 0)p2++;
        cnt[p2-p1+2]+=1;
        p1 = p2+1;
    }
}

signed main(){
    Orz;
    cin>>n;
    p.assign(n,{0,0});
    cnt.resize(n+1,0);
    rep(i,0,n-1)cin>>p[i].x>>p[i].y;
    rep(i,0,n-1)solve(p[i]);
    int ans = (n*(n-1)*(n-2))/6;
    rep(i,3,n)ans-=(cnt[i]*(i-1)*(i-2))/6;
    cout<<ans<<endl;
}