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

<image>

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>
using namespace std;
template <class T>
inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmin(T &a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
struct query {
  int index;
  long double s, r;
  query(int _index, int _s, int _r) {
    index = _index;
    s = _s;
    r = _r;
  }
  query() {
    index = 0;
    s = 0;
    r = 0;
  }
};
const long double EPS = 1e-10;
bool operator>(query a, query b) {
  if (abs(a.s - b.s) > EPS) return a.s > b.s;
  if (abs(a.r - b.r) > EPS) return a.r > b.r;
  return a.index > b.index;
}
bool operator<(query a, query b) {
  if (abs(a.s - b.s) > EPS) return a.s < b.s;
  if (abs(a.r - b.r) > EPS) return a.r < b.r;
  return a.index < b.index;
}
int N;
bool win(query a, query b, long double R) {
  long double ascore = 1 / a.s + R / a.r;
  long double bscore = 1 / b.s + R / b.r;
  return ascore - bscore <= EPS;
}
long double getR(query a, query b, long double minimum) {
  long double ok = 1e18;
  long double ng = minimum;
  for (int timer = 0; timer <= 500; timer++) {
    long double mid = (ok + ng) / 2.0;
    if (win(a, b, mid))
      ok = mid;
    else
      ng = mid;
  }
  return ok;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> N;
  vector<query> input(N);
  for (int i = 1; i <= N; i++) {
    int r, s;
    cin >> s >> r;
    input[i - 1].index = i;
    input[i - 1].s = s;
    input[i - 1].r = r;
  }
  sort(input.begin(), input.end(), greater<query>());
  vector<query> candidate;
  long double rmax = 0.0;
  long double rmaxbefore = 0.0;
  for (int i = 0; i < input.size(); i++) {
    if (i >= 1 && input[i - 1].s - input[i].s > EPS) rmaxbefore = rmax;
    if (rmax - input[i].r >= EPS) continue;
    chmax(rmax, input[i].r);
    if (!(input[i].r - rmaxbefore >= EPS)) continue;
    candidate.push_back(input[i]);
  }
  for (int i = 0; i < candidate.size(); i++) cerr << candidate[i].index << endl;
  vector<pair<query, long double> > v;
  v.push_back({candidate[0], 0.0});
  for (int i = 1; i < candidate.size(); i++) {
    while (!v.empty()) {
      if (!win(v.back().first, candidate[i], v.back().second)) {
        v.pop_back();
      } else {
        break;
      }
    }
    long double minimum = getR(candidate[i], v.back().first, v.back().second);
    cerr << candidate[i].index << " wins against " << v.back().first.index
         << " with R=" << minimum << endl;
    v.push_back({candidate[i], minimum});
  }
  vector<int> ans;
  for (int i = 0; i < v.size(); i++) {
    ans.push_back(v[i].first.index);
  }
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}
