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 
5
10 50
10 50
10 50
10 50
10 50
 2
10000 10000
10000 10000
 3
10000 10000
10000 10000
10000 10000
 3
1000 3000
1500 1500
3000 1000
 43
75 16
70 7
75 16
54 2
70 7
54 2
70 7
54 2
70 7
54 2
54 2
70 7
70 7
54 2
70 7
75 16
70 7
54 2
54 2
54 2
54 2
54 2
54 2
75 16
70 7
70 7
54 2
70 7
75 16
54 2
75 16
75 16
54 2
54 2
70 7
70 7
75 16
70 7
70 7
70 7
75 16
75 16
54 2
 1
1 1
 2
1 1
1 1
 44
78 52
75 16
70 7
75 16
54 2
70 7
54 2
70 7
54 2
70 7
54 2
54 2
70 7
70 7
54 2
70 7
75 16
70 7
54 2
54 2
54 2
54 2
54 2
54 2
75 16
70 7
70 7
54 2
70 7
75 16
54 2
75 16
75 16
54 2
54 2
70 7
70 7
75 16
70 7
70 7
70 7
75 16
75 16
54 2
 3
1000 3000
1499 1499
3000 1000
 5
50 50
49 50
50 49
49 49
50 1
 18
82 38
33 69
33 69
33 69
33 69
33 69
82 38
33 69
33 69
33 69
82 38
33 69
33 69
82 38
82 38
33 69
33 69
33 69
 1
10000 9999
 3
1000 3000
1501 1501
3000 1000
 2
1 2
2 1

1 2 3 4 5
 1 2
 1 2 3
 1 2 3
 1 3 16 24 29 31 32 37 41 42
 1
 1 2
 1
 1 3
 1
 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
 1
 1 2 3
 1 2 

#include <bits/stdc++.h>
using namespace std;
int n, a[200500], b[200500];
map<int, int> max_a, max_b;
set<pair<int, int> > alive;
inline void updmax(int &to, int fr) { to = max(to, fr); }
inline bool bigger(pair<long long, long long> fi,
                   pair<long long, long long> se) {
  long long high = fi.first * se.second - se.first * fi.second;
  long long low = fi.second * se.second;
  return !((high > 0) ^ (low > 0));
}
int main() {
  cin.sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i] >> b[i];
    updmax(max_a[b[i]], a[i]);
    updmax(max_b[a[i]], b[i]);
  }
  vector<pair<int, int> > proc;
  for (int i = 0; i < n; i++) {
    if (max_a[b[i]] != a[i] or max_b[a[i]] != b[i]) continue;
    proc.emplace_back(make_pair(a[i], b[i]));
  }
  sort(proc.begin(), proc.end());
  proc.erase(unique(proc.begin(), proc.end()), proc.end());
  int sz = proc.size();
  for (int i = 0; i < sz; i++) {
    bool good = true;
    pair<long long, long long> mn(-1 * 12312333ll, 1);
    pair<long long, long long> mx(1 * 12312333ll, 1);
    for (int j = 0; good and j < sz; j++) {
      if (j == i) continue;
      long long up = (proc[j].second - proc[i].second) * proc[j].first;
      long long down = (proc[i].first - proc[j].first) * proc[j].second;
      if (proc[i].first < proc[j].first and proc[i].second < proc[j].second)
        good = false;
      pair<long long, long long> cur(up, down);
      if (proc[i].first - proc[j].first > 0) {
        if (bigger(cur, mn)) mn = cur;
      } else {
        if (bigger(mx, cur)) mx = cur;
      }
    }
    good &= bigger(mx, mn) or (mx.first * mn.second == mx.second * mn.first);
    if (good) alive.insert(proc[i]);
  }
  for (int i = 0; i < n; i++) {
    if (alive.count(make_pair(a[i], b[i])))
      if (n < -200)
        cout << i + 1 << " ";
      else
        cout << a[i] << " " << b[i] << endl;
  }
  return 0;
}
