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;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vll = vector<ll>;
template <typename T>
T maximize(T& x, const T& y) {
  if (y > x) x = y;
  return x;
}
template <typename T>
T minimize(T& x, const T& y) {
  if (y < x) x = y;
  return x;
}
template <class T, class U>
std::ostream& operator<<(std::ostream& os, const pair<T, U>& p) {
  return os << "{" << p.first << "," << p.second << "}";
}
const double eps = 1e-9;
bool lt(double a, double b) { return a < b - eps; }
bool eq(double a, double b) { return !lt(a, b) && !lt(b, a); }
bool Q;
struct Line {
  mutable double k, m, p;
  bool operator<(const Line& o) const { return Q ? p < o.p : k < o.k; }
};
struct LineContainer : multiset<Line> {
  const double inf = 1 / .0;
  double div(double a, double b) { return a / b; }
  bool isect(iterator x, iterator y) {
    if (y == end()) {
      x->p = inf;
      return false;
    }
    if (eq(x->k, y->k))
      x->p = x->m > y->m ? inf : -inf;
    else
      x->p = div(y->m - x->m, x->k - y->k);
    return !lt(x->p, y->p);
  }
  void add_min(double k, double m) { add(-k, -m); }

 private:
  void add(double k, double m) {
    auto z = insert({k, m, 0}), y = z++, x = y;
    while (isect(y, z)) z = erase(z);
    if (x != begin() && isect(--x, y)) isect(x, y = erase(y));
    while ((y = x) != begin() && !lt((--x)->p, y->p)) isect(x, erase(y));
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cerr.tie(NULL);
  int n;
  cin >> n;
  LineContainer lines;
  map<pii, vi> indices;
  for (auto i = (1); i <= (n); ++i) {
    int s, r;
    cin >> s >> r;
    pii p = make_pair(s, r);
    if (!indices.count(p)) lines.add_min(1.0 / s, 1.0 / r);
    indices[p].push_back(i);
  }
  vi result;
  for (auto& line : lines)
    if (lt(.0, line.p)) {
      int s = round(1.0 / -line.k);
      int r = round(1.0 / -line.m);
      pii p = make_pair(s, r);
      vi ids = indices[p];
      result.insert(result.end(), (ids).begin(), (ids).end());
    }
  sort((result).begin(), (result).end());
  for (auto& i : result) cout << i << " ";
  cout << endl;
  return 0;
}
