#include <bits/stdc++.h>
#ifdef DEBUG
#include <debug.cpp>
#else
#define debug(...)
#endif

using namespace std;
using ll = long long;
using lf = double;

constexpr double inf = 1e100;

// 向量
struct vec {
    static bool cmp(const vec &a, const vec &b) { return tie(a.x, a.y) < tie(b.x, b.y); }

    ll x, y;
    vec() : x(0), y(0) {}
    vec(ll _x, ll _y) : x(_x), y(_y) {}

    vec rotleft() const { return {-y, x}; }
    vec rotright() const { return {y, -x}; }

    // 模
    ll len2() const { return x * x + y * y; }
    double len() const { return sqrt(x * x + y * y); }

    // 是否在上半轴
    bool up() const { return y > 0 || y == 0 && x >= 0; }

    bool operator==(const vec &b) const { return tie(x, y) == tie(b.x, b.y); }
    // 极角排序
    bool operator<(const vec &b) const {
        return cmp(*this, b);
        if (up() != b.up()) return up() > b.up();
        ll tmp = (*this) ^ b;
        return tmp ? tmp > 0 : cmp(*this, b);
    }

    vec operator+(const vec &b) const { return {x + b.x, y + b.y}; }
    vec operator-() const { return {-x, -y}; }
    vec operator-(const vec &b) const { return -b + (*this); }
    vec operator*(ll b) const { return {x * b, y * b}; }
    ll operator*(const vec &b) const { return x * b.x + y * b.y; }

    // 叉积 结果大于0，a到b为逆时针，小于0，a到b顺时针,
    // 等于0共线，可能同向或反向，结果绝对值表示 a b 形成的平行四边行的面积
    ll operator^(const vec &b) const { return x * b.y - y * b.x; }

    friend istream &operator>>(istream &in, vec &data) {
        in >> data.x >> data.y;
        return in;
    }
    friend ostream &operator<<(ostream &out, const vec &data) {
        out << data.x << " " << data.y;
        return out;
    }
};

lf angle(const vec &a, const vec &b) { return atan2(abs(a ^ b), a * b); }

ll cross(const vec &a, const vec &b, const vec &c) { return (a - c) ^ (b - c); }

// 多边形的面积a
double polygon_area(vector<vec> &p) {
    ll area = 0;
    for (int i = 1; i < p.size(); i++) area += p[i - 1] ^ p[i];
    area += p.back() ^ p[0];
    return abs(area / 2.0);
}

// 多边形的周长
double polygon_len(vector<vec> &p) {
    double len = 0;
    for (int i = 1; i < p.size(); i++) len += (p[i - 1] - p[i]).len();
    len += (p.back() - p[0]).len();
    return len;
}

// 以整点为顶点的线段上的整点个数
ll count(const vec &a, const vec &b) {
    vec c = a - b;
    return gcd(abs(c.x), abs(c.y)) + 1;
}

// 以整点为顶点的多边形边上整点个数
ll count(vector<vec> &p) {
    ll cnt = 0;
    for (int i = 1; i < p.size(); i++) cnt += count(p[i - 1], p[i]);
    cnt += count(p.back(), p[0]);
    return cnt - p.size();
}

// 判断点是否在凸包内，凸包必须为逆时针顺序
bool in_polygon(const vec &a, vector<vec> &p) {
    int n = p.size();
    if (n == 0) return 0;
    if (n == 1) return a == p[0];
    if (n == 2) return cross(a, p[1], p[0]) == 0 && (p[0] - a) * (p[1] - a) <= 0;
    if (cross(a, p[1], p[0]) > 0 || cross(p.back(), a, p[0]) > 0) return 0;
    auto cmp = [&](vec &x, const vec &y) { return ((x - p[0]) ^ y) >= 0; };
    int i = lower_bound(p.begin() + 2, p.end() - 1, a - p[0], cmp) - p.begin() - 1;
    return cross(p[(i + 1) % n], a, p[i]) >= 0;
}

// 凸包直径的两个端点
auto polygon_dia(vector<vec> &p) {
    int n = p.size();
    array<vec, 2> res{};
    if (n == 1) return res;
    if (n == 2) return res = {p[0], p[1]};
    ll mx = 0;
    for (int i = 0, j = 2; i < n; i++) {
        while (abs(cross(p[i], p[(i + 1) % n], p[j])) <=
               abs(cross(p[i], p[(i + 1) % n], p[(j + 1) % n])))
            j = (j + 1) % n;
        ll tmp = (p[i] - p[j]).len2();
        if (tmp > mx) {
            mx = tmp;
            res = {p[i], p[j]};
        }
        tmp = (p[(i + 1) % n] - p[j]).len2();
        if (tmp > mx) {
            mx = tmp;
            res = {p[(i + 1) % n], p[j]};
        }
    }
    return res;
}

// 凸包
auto convex_hull(vector<vec> &p) {
    sort(p.begin(), p.end(), vec::cmp);
    int n = p.size();
    vector sta(n + 1, 0);
    vector v(n, false);
    int tp = -1;
    sta[++tp] = 0;
    auto update = [&](int lim, int i) {
        while (tp > lim && cross(p[i], p[sta[tp]], p[sta[tp - 1]]) >= 0) v[sta[tp--]] = 0;
        sta[++tp] = i;
        v[i] = 1;
    };
    for (int i = 1; i < n; i++) update(0, i);
    int cnt = tp;
    for (int i = n - 1; i >= 0; i--) {
        if (v[i]) continue;
        update(cnt, i);
    }
    vector<vec> res(tp);
    for (int i = 0; i < tp; i++) res[i] = p[sta[i]];
    return res;
}

// 闵可夫斯基和，两个点集的和构成一个凸包
auto minkowski(vector<vec> &a, vector<vec> &b) {
    rotate(a.begin(), min_element(a.begin(), a.end(), vec::cmp), a.end());
    rotate(b.begin(), min_element(b.begin(), b.end(), vec::cmp), b.end());
    int n = a.size(), m = b.size();
    vector<vec> c{a[0] + b[0]};
    c.reserve(n + m);
    int i = 0, j = 0;
    while (i < n && j < m) {
        vec x = a[(i + 1) % n] - a[i];
        vec y = b[(j + 1) % m] - b[j];
        c.push_back(c.back() + ((x ^ y) >= 0 ? (i++, x) : (j++, y)));
    }
    while (i + 1 < n) {
        c.push_back(c.back() + a[(i + 1) % n] - a[i]);
        i++;
    }
    while (j + 1 < m) {
        c.push_back(c.back() + b[(j + 1) % m] - b[j]);
        j++;
    }
    return c;
}

// 过凸多边形外一点求凸多边形的切线，返回切点下标
auto tangent(const vec &a, vector<vec> &p) {
    int n = p.size();
    int l = -1, r = -1;
    for (int i = 0; i < n; i++) {
        ll tmp1 = cross(p[i], p[(i - 1 + n) % n], a);
        ll tmp2 = cross(p[i], p[(i + 1) % n], a);
        if (l == -1 && tmp1 <= 0 && tmp2 <= 0) l = i;
        else if (r == -1 && tmp1 >= 0 && tmp2 >= 0) r = i;
    }
    return array{l, r};
}

// 直线
struct line {
    vec p, d;
    line() {}
    line(const vec &a, const vec &b) : p(a), d(b - a) {}
};

// 点到直线距离
double dis(const vec &a, const line &b) { return abs((b.p - a) ^ (b.p + b.d - a)) / b.d.len(); }

// 点在直线哪边，大于0在左边，等于0在线上，小于0在右边
ll side_line(const vec &a, const line &b) { return b.d ^ (a - b.p); }

// 两直线是否垂直
bool perpen(const line &a, const line &b) { return a.d * b.d == 0; }

// 两直线是否平行
bool parallel(const line &a, const line &b) { return (a.d ^ b.d) == 0; }

// 点的垂线是否与线段有交点
bool perpen(const vec &a, const line &b) {
    vec p(-b.d.y, b.d.x);
    bool cross1 = (p ^ (b.p - a)) > 0;
    bool cross2 = (p ^ (b.p + b.d - a)) > 0;
    return cross1 != cross2;
}

// 点到线段距离
double dis_seg(const vec &a, const line &b) {
    if (perpen(a, b)) return dis(a, b);
    return min((b.p - a).len(), (b.p + b.d - a).len());
}

// 点到凸包距离
double dis(const vec &a, vector<vec> &p) {
    double res = inf;
    for (int i = 1; i < p.size(); i++) res = min(dis_seg(a, line(p[i - 1], p[i])), res);
    res = min(dis_seg(a, line(p.back(), p[0] - p.back())), res);
    return res;
}

// 两直线交点
vec intersection(ll A, ll B, ll C, ll D, ll E, ll F) {
    return {(B * F - C * E) / (A * E - B * D), (C * D - A * F) / (A * E - B * D)};
}

// 两直线交点
vec intersection(const line &a, const line &b) {
    return intersection(a.d.y, -a.d.x, a.d.x * a.p.y - a.d.y * a.p.x, b.d.y, -b.d.x,
                        b.d.x * b.p.y - b.d.y * b.p.x);
}

void solve() {
    int n;
    cin >> n;
    vector<vec> p(n);
    for (auto &i : p) cin >> i;

    auto convex = convex_hull(p);
    debug(convex);
    set<vec> S(convex.begin(), convex.end());
    convex.push_back(convex[0]);
    vector<vec> p2;
    for (auto &i : p) {
        if (S.count(i)) continue;
        p2.push_back(i);
    }
    int ans = 1;
    for (auto &o : p2) {
        vector<vec> p3;
        for (auto &i : p2) {
            if (o == i) continue;
            p3.push_back(i - o);
        }
        lf mn = 1e100;
        int pos = 0;
        for (auto &i : p3) {
            lf tmp = angle(convex[0] - o, i);
            if (tmp < mn) {
                mn = tmp;
                pos = &i - &p3[0];
            }
        }
        rotate(p3.begin(), p3.begin() + pos, p3.end());
        for (int i = 1, j = 0; i < convex.size(); i++) {
            bool ok = 1;
            while (j < p3.size() && ((convex[i - 1] - o) ^ p3[j]) >= 0 &&
                   (p3[j] ^ (convex[i] - o)) >= 0) {
                if (((convex[i - 1] - o) ^ p3[j]) > 0 && (p3[j] ^ (convex[i] - o)) > 0) ok = 0;
                j++;
            }
            debug(o, i, ok);
            ans += ok;
        }
    }
    cout << ans << "\n";
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t = 1;
    // cin >> t;
    while (t--) solve();
}