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;
template <typename G1, typename G2 = G1, typename G3 = G1>
struct triple {
  G1 first;
  G2 second;
  G3 T;
};
const long double eps = 1e-12;
int sign(long double x) { return x < -eps ? -1 : x > eps; }
int dblcmp(long double x, long double y) { return sign(x - y); }
long double dot(complex<long double> a, complex<long double> b) {
  return real(conj(a) * b);
}
long double cross(complex<long double> a, complex<long double> b) {
  return imag(conj(a) * b);
}
struct Point {
  complex<long double> p;
  int id;
  long double operator*(const Point o) const { return cross(this->p, o.p); }
  Point operator-(const Point o) const { return {p - o.p}; }
  bool operator<(const Point o) const {
    if (dblcmp(p.real(), o.p.real()) == 0)
      return dblcmp(p.imag(), o.p.imag()) < 0;
    return dblcmp(p.real(), o.p.real()) < 0;
  }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<vector<int>> mp1(n);
  map<pair<long long, long long>, int> mp2;
  vector<Point> v;
  int c = 0;
  for (int i = 0; i < n; i++) {
    int x, y;
    cin >> x >> y;
    if (mp2.count({x, y}) == 0) {
      v.push_back({{10000.0 / x, 10000.0 / y}, c});
      mp2[{x, y}] = c++;
    }
    mp1[mp2[{x, y}]].push_back(i + 1);
  }
  sort(v.begin(), v.end());
  vector<Point> hull;
  for (int i = 0; i < c; i++) {
    while (hull.size() > 1 &&
           dblcmp((hull.end()[-1] - hull.end()[-2]) * (v[i] - hull.end()[-1]),
                  0) < 0)
      hull.pop_back();
    hull.push_back(v[i]);
  }
  while (hull.size() > 1 &&
         dblcmp(hull.end()[-2].p.imag(), hull.end()[-1].p.imag()) <= 0)
    hull.pop_back();
  vector<int> ans;
  for (auto &i : hull)
    for (auto j : mp1[i.id]) ans.push_back(j);
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); i++)
    cout << ans[i] << " \n"[i == ans.size() - 1];
  return 0;
}
