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;
typedef double ld;
struct info {
  complex<ld> p;
  int id;
  bool operator<(const info &o) const {
    if (p.real() != o.p.real()) return p.real() < o.p.real();
    return p.imag() < o.p.imag();
  }
  bool operator==(const info &o) const { return p == o.p; }
};
ld cross(complex<ld> a, complex<ld> b) { return imag(conj(a) * b); }
ld cross(info a, info b, info c) { return cross(b.p - a.p, c.p - a.p); }
const ld eps = 1e-22;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int n;
  cin >> n;
  vector<info> a(n);
  for (auto &i : a) {
    int x, y;
    cin >> x >> y;
    i.p = {(ld)1.0 / x, (ld)1.0 / y};
    i.id = &i - &a[0];
  }
  map<info, vector<int>> mp;
  for (auto i : a) mp[i].push_back(i.id);
  vector<info> ans;
  sort(a.begin(), a.end());
  a.erase(unique(a.begin(), a.end()), a.end());
  for (auto i : a) {
    while (ans.size() >= 2 && cross(ans.end()[-2], ans.end()[-1], i) < -eps)
      ans.pop_back();
    ans.push_back(i);
  }
  while (ans.size() >= 2 && (ans.end()[-1].p - ans.end()[-2].p).real() < eps)
    ans.pop_back();
  vector<int> pos;
  for (auto i : ans)
    for (auto j : mp[i]) pos.push_back(j);
  sort(pos.begin(), pos.end());
  for (auto i : pos) cout << i + 1 << " ";
  return 0;
}
