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 
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;
const int MAXN = 1001 * 201;
pair<int, pair<int, vector<int> > > a[MAXN];
vector<pair<int, pair<int, int> > > v;
vector<int> ans;
double mn[MAXN] = {}, mx[MAXN] = {};
map<pair<int, int>, int> m;
int main() {
  ios::sync_with_stdio(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i].first >> a[i].second.first;
    if (!m[{a[i].first, a[i].second.first}])
      m[{a[i].first, a[i].second.first}] = i + 1;
    a[m[{a[i].first, a[i].second.first}] - 1].second.second.push_back(i);
  }
  sort(a, a + n);
  reverse(a, a + n);
  int mxx = -1;
  for (int i = 0; i < n; i++) {
    if (a[i].second.first > mxx) {
      mxx = a[i].second.first;
      for (auto u : a[i].second.second) v.push_back({a[i].first, {mxx, u}});
    }
  }
  for (int i = 0; i < n; i++) {
    mx[i] = 1e10;
    mn[i] = -100000;
  }
  n = v.size();
  for (int i = 0; i < n; i++) {
    double ri = v[i].first, si = v[i].second.first;
    for (int j = i + 1; j < n; j++) {
      double rj = v[j].first, sj = v[j].second.first;
      if (ri == rj && si == sj) continue;
      double x = (ri * rj * (si - sj)) / (si * sj * (rj - ri));
      mx[j] = min(mx[j], x);
      mn[i] = max(mn[i], x);
    }
  }
  cout << setprecision(10);
  for (int i = 0; i < n; i++)
    if (mn[i] <= mx[i]) ans.push_back(v[i].second.second);
  sort(ans.begin(), ans.end());
  for (auto u : ans) cout << u + 1 << ' ';
}
