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;
pair<int, pair<int, int> > v[200002];
bool vis[200002];
struct Hull {
  deque<pair<double, double> > v;
  deque<int> id;
  double cross(pair<double, double> l1, pair<double, double> l2) {
    return double(l2.second - l1.second) / double(l1.first - l2.first);
  }
  void Insert(double m, double b, int _id) {
    pair<double, double> line = make_pair(m, b);
    while ((int)v.size() > 1 &&
           cross(line, v.back()) < cross(v.back(), v[v.size() - 2]) - 1e9)
      v.pop_back(), id.pop_back();
    v.push_back(line);
    id.push_back(_id);
  }
  void solve() {
    sort(id.begin(), id.end());
    for (int i = 0; i < (int)id.size(); i++) printf("%d ", id[i]);
    printf("\n");
  }
};
int main() {
  int n, rr;
  scanf("%d", &n);
  for (int i = 0; i < n; i++)
    scanf("%d%d", &v[i].first, &v[i].second.first), v[i].second.second = i + 1;
  sort(v, v + n);
  rr = n - 1;
  Hull G;
  vis[rr] = 1;
  for (int i = n - 2; i >= 0; i--)
    if (v[i].second.first > v[rr].second.first ||
        (v[i].first == v[rr].first && v[i].second.first == v[rr].second.first))
      vis[i] = 1, rr = i;
  for (int i = 0; i < n; i++) {
    if (vis[i]) {
      G.Insert(double(v[i].second.first - v[i].first) /
                   double(v[i].second.first * v[i].first),
               1e6 / double(v[i].second.first), v[i].second.second);
    }
  }
  G.solve();
}
