#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using Point = pair<ll, ll>;

// return cross product of (b - a) x (c - a)
ll cross(const Point &a, const Point &b, const Point &c) {
    return (b.first - a.first) * (c.second - a.second)
         - (b.second - a.second) * (c.first - a.first);
}

vector<Point> convex_hull(vector<Point> pts) {
    int n = pts.size();
    if (n < 3) return pts;

    // 1) find pivot: smallest x (and then y) point
    int pivot = 0;
    for (int i = 1; i < n; i++) {
        if (pts[i].first < pts[pivot].first ||
           (pts[i].first == pts[pivot].first && pts[i].second < pts[pivot].second))
            pivot = i;
    }
    swap(pts[0], pts[pivot]);
    Point P0 = pts[0];

    // 2) sort by polar angle around P0, tie-break by distance
    sort(pts.begin() + 1, pts.end(), [&](const Point &a, const Point &b) {
        ll c = cross(P0, a, b);
        if (c != 0) return c > 0;
        // if collinear, put closer point first
        ll da = (a.first - P0.first)*(a.first - P0.first)
              + (a.second - P0.second)*(a.second - P0.second);
        ll db = (b.first - P0.first)*(b.first - P0.first)
              + (b.second - P0.second)*(b.second - P0.second);
        return da < db;
    });

    // 3) Graham scan
    vector<Point> hull;
    hull.push_back(pts[0]);
    hull.push_back(pts[1]);
    for (int i = 2; i < n; i++) {
        while (hull.size() >= 2 &&
               cross(hull[hull.size()-2], hull[hull.size()-1], pts[i]) <= 0) {
            // remove non-left turn (<=0 removes collinear as well)
            hull.pop_back();
        }
        hull.push_back(pts[i]);
    }
    return hull;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int N;
    cin >> N;
    vector<Point> pts(N);
    for (int i = 0; i < N; i++) {
        cin >> pts[i].first >> pts[i].second;
    }

    // compute convex hull
    vector<Point> hull = convex_hull(pts);

    // if hull does not include all points, or N<3
    if ((int)hull.size() != N) {
        cout << "No\n";
        return 0;
    }

    // check no three consecutive points are collinear
    // wrap around: treat hull as circular
    for (int i = 0; i < N; i++) {
        Point a = hull[i];
        Point b = hull[(i+1)%N];
        Point c = hull[(i+2)%N];
        if (cross(a, b, c) == 0) {
            cout << "No\n";
            return 0;
        }
    }

    cout << "Yes\n";
    return 0;
}