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;
const double eps = 1e-8;
const int N = 2e5 + 5;
bool mark[N];
int n, id[N];
complex<double> a[N];
char direction(int p1, int p2, int p3) {
  double x = (a[p2] - a[p1]).real() * (a[p3] - a[p1]).imag() -
             (a[p2] - a[p1]).imag() * (a[p3] - a[p1]).real();
  return abs(x) < eps ? 'S' : x < 0 ? 'R' : 'L';
}
void read_input() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x, y;
    cin >> x >> y;
    a[i] = complex<double>(1. / x, 1. / y);
  }
}
void solve() {
  iota(id, id + n, 0);
  sort(id, id + n, [](int i, int j) {
    return make_pair(a[i].real(), a[i].imag()) <
           make_pair(a[j].real(), a[j].imag());
  });
  int mn = id[0];
  for (int i = 0; i < n; i++)
    if (make_pair(a[i].imag(), a[i].real()) <
        make_pair(a[mn].imag(), a[mn].real()))
      mn = i;
  vector<int> A = {mn};
  for (int i = 0; id[i] ^ mn; A.push_back(id[i++]))
    while (A.size() > 1 && direction(A[A.size() - 2], A.back(), id[i]) == 'R')
      A.pop_back();
  for (auto x : A) mark[x] = true;
}
void write_output() {
  for (int i = 0; i < n; i++)
    if (mark[i]) cout << i + 1 << ' ';
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  read_input(), solve(), write_output();
  return 0;
}
