#include <bits/stdc++.h>
using namespace std;

#include <atcoder/all>
using namespace atcoder;

using ll = long long;
using mint = modint998244353;

int N;
ll X[2 << 17];
ll Y[2 << 17];

ll cross(int i, int j, int k) {
  ll xi = X[i] - X[j];
  ll yi = Y[i] - Y[j];
  ll xk = X[k] - X[j];
  ll yk = Y[k] - Y[j];
  return xk * yi - yk * xi;
}

ll dist(int i, int j) {
  ll xi = X[i] - X[j];
  ll yi = Y[i] - Y[j];
  return xi * xi + yi * yi;
}

pair<ll, ll> tan(int i, int j) {
  ll dx = X[j] - X[i];
  ll dy = Y[j] - Y[i];
  ll g = gcd(dx, dy);
  ll sgn = dx >= 0 ? 1 : -1;
  return make_pair(dx / g * sgn, dy / g * sgn);
}

int find_pivot() {
  vector<int> P(N);
  iota(P.begin(), P.end(), 0);
  auto it = min_element(P.begin(), P.end(), [&](int i, int j) {
    return X[i] != X[j] ? X[i] < X[j] : Y[i] < Y[j];
  });
  return it - P.begin();
}

int main() {
  cin >> N;
  for(int i = 0; i < N; i++) {
    cin >> X[i] >> Y[i];
  }
  const int k = find_pivot();
  set<pair<ll, ll>> st_tan;
  for(int i = 0; i < N; i++) {
    if(i == k) continue;
    st_tan.insert(tan(k, i));
  }
  if(ssize(st_tan) < N - 1) {
    cout << "No" << endl;
    return 0;
  }
  vector<int> P(N);
  iota(P.begin(), P.end(), 0);
  P.erase(P.begin() + k);
  sort(P.begin(), P.end(), [&](int i, int j) {
    return cross(k, i, j) > 0;
  });
  P.push_back(k);
  for(int x = 0; x < N; x++) {
    int y = (x + 1) % N;
    int z = (x + 2) % N;
    if(cross(P[x], P[y], P[z]) <= 0) {
      cout << "No" << endl;
      return 0;
    }
  }
  cout << "Yes" << endl;
}