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;
int prodvec(pair<int, int> p1, pair<int, int> p2) {
  return p1.first * p2.second - p1.second * p2.first;
}
int modulo(pair<int, int> p) { return p.first * p.first + p.second * p.second; }
pair<int, int> operator-(pair<int, int> p1, pair<int, int> p2) {
  p1.first -= p2.first;
  p1.second -= p2.second;
  return p1;
}
int m;
pair<int, int> p[1000000];
bool compara(pair<int, int> p1, pair<int, int> p2) {
  int pv = prodvec(p1 - p[0], p2 - p[0]);
  if (pv > 0) return true;
  if (pv < 0) return false;
  return modulo(p1) < modulo(p2);
}
map<pair<int, int>, vector<int> > mapa;
vector<int> sol;
void anyade(pair<int, int> po) {
  vector<int> &v = mapa[po];
  for (int i = 0; i < int(v.size()); i++) sol.push_back(v[i]);
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  int maxfirst = 0, maxsecond = 0;
  for (int i = 1; i <= n; i++) {
    pair<int, int> po;
    cin >> po.first >> po.second;
    mapa[po].push_back(i);
    maxfirst = max(maxfirst, po.first);
    maxsecond = max(maxsecond, po.second);
  }
  m = 0;
  p[m++] = pair<int, int>(0, 0);
  p[m++] = pair<int, int>(maxfirst, 0);
  p[m++] = pair<int, int>(maxsecond, 0);
  for (map<pair<int, int>, vector<int> >::iterator it = mapa.begin();
       it != mapa.end(); it++)
    p[m++] = it->first;
  sort(p + 1, p + m, compara);
  vector<pair<int, int> > convex;
  convex.push_back(p[0]);
  for (int i = 1; i < m; i++) {
    pair<int, int> &po = p[i];
    while (
        int(convex.size()) >= 2 and
        prodvec(convex[int(convex.size()) - 1] - convex[int(convex.size()) - 2],
                po - convex[int(convex.size()) - 1]) < 0)
      convex.pop_back();
    convex.push_back(po);
  }
  for (int i = 0; i < int(convex.size()); i++) anyade(convex[i]);
  sort(sol.begin(), sol.end());
  for (int i = 0; i < int(sol.size()); i++) {
    if (i > 0) cout << " ";
    cout << sol[i];
  }
  cout << endl;
}
