Tavas is a cheerleader in the new sports competition named "Pashmaks".

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This competition consists of two part: swimming and then running. People will immediately start running R meters after they finished swimming exactly S meters. A winner is a such person that nobody else finishes running before him/her (there may be more than one winner).

Before the match starts, Tavas knows that there are n competitors registered for the match. Also, he knows that i-th person's swimming speed is si meters per second and his/her running speed is ri meters per second. Unfortunately, he doesn't know the values of R and S, but he knows that they are real numbers greater than 0.

As a cheerleader, Tavas wants to know who to cheer up. So, he wants to know all people that might win. We consider a competitor might win if and only if there are some values of R and S such that with these values, (s)he will be a winner.

Tavas isn't really familiar with programming, so he asked you to help him.

Input

The first line of input contains a single integer n (1 ≤ n ≤ 2 × 105).

The next n lines contain the details of competitors. i-th line contains two integers si and ri (1 ≤ si, ri ≤ 104).

Output

In the first and the only line of output, print a sequence of numbers of possible winners in increasing order.

Examples

Input

3
1 3
2 2
3 1


Output

1 2 3 


Input

3
1 2
1 1
2 1


Output

1 3 
#include <bits/stdc++.h>
const int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};
using namespace std;
double eps = 1e-22;
double add(double a, double b) {
  if (abs(a + b) < eps * (abs(a) + abs(b))) return 0;
  return a + b;
}
bool equal(double a, double b) { return add(a, -b) == 0; }
struct P {
  double x, y;
  P() {}
  P(double x, double y) : x(x), y(y) {}
  P operator+(P p) const { return P(add(x, p.x), add(y, p.y)); }
  P operator-(P p) const { return P(add(x, -p.x), add(y, -p.y)); }
  P operator*(double d) const { return P(x * d, y * d); }
  double dot(P p) const { return add(x * p.x, y * p.y); }
  double det(P p) const { return add(x * p.y, -y * p.x); }
  double dist(P p) const {
    return (x - p.x) * (x - p.x) + (y - p.y) * (y - p.y);
  }
  bool operator<(const P& rhs) const {
    if (x != rhs.x) return x < rhs.x;
    return y < rhs.y;
  }
  bool operator==(const P& rhs) const {
    return equal(x, rhs.x) && equal(y, rhs.y);
  }
};
bool on_seg(P p1, P p2, P q) {
  return (p1 - q).det(p2 - q) == 0 && (p1 - q).dot(p2 - q) <= 0;
}
P intersection(P p1, P p2, P q1, P q2) {
  return p1 + (p2 - p1) * ((q2 - q1).det(q1 - p1) / (q2 - q1).det(p2 - p1));
}
vector<P> convex_hull(vector<P> ps) {
  int n = ps.size();
  sort(ps.begin(), ps.end());
  int k = 0;
  vector<P> qs(n * 2);
  for (int i = 0; i < n; i++) {
    while (k > 1 && (qs[k - 1] - qs[k - 2]).det(ps[i] - qs[k - 1]) <= 0) k--;
    qs[k++] = ps[i];
  }
  qs.resize(k);
  return qs;
  for (int i = n - 2, t = k; i >= 0; i--) {
    while (k > t && (qs[k - 1] - qs[k - 2]).det(ps[i] - qs[k - 1]) <= 0) k--;
    qs[k++] = ps[i];
  }
  qs.resize(k - 1);
  return qs;
}
int ma[11111];
int s[211111], r[211111];
vector<int> memo[11111];
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<P> ps(n);
  map<P, vector<int> > m;
  for (int i = 0; i < n; i++) {
    cin >> s[i] >> r[i];
    ps[i] = P(1. / s[i], 1. / r[i]);
    m[ps[i]].push_back(i);
    ma[s[i]] = max(r[i], ma[s[i]]);
  }
  for (int i = 0; i < n; i++) {
    if (r[i] == ma[s[i]]) memo[s[i]].push_back(i + 1);
  }
  if (n <= 2) {
    for (int i = 0; i < n; i++) {
      cout << i + 1 << " ";
    }
    cout << endl;
    return 0;
  }
  auto qs = convex_hull(ps);
  sort(qs.begin(), qs.end());
  int y = -1;
  vector<int> ans;
  for (int i = 10000; i >= 1; i--) {
    if (memo[i].size() == 0) continue;
    if (y != -1 && ma[y] >= ma[i]) continue;
    P p = P(1. / i, 1. / ma[i]);
    if (find(qs.begin(), qs.end(), p) != qs.end()) {
      for (int el : memo[i]) {
        ans.push_back(el);
      }
    }
    y = i;
  }
  sort(ans.begin(), ans.end());
  for (int el : ans) {
    cout << el << " ";
  }
  cout << endl;
  return 0;
}
