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>
const double eps = 1e-12;
using namespace std;
struct pdd {
  double x, y;
  int id;
};
double cross(const pdd &a, const pdd &b) { return a.x * b.y - a.y * b.x; }
double dot(const pdd &a, const pdd &b) { return a.x * b.x + a.y * b.y; }
pdd sub(const pdd &a, const pdd &b) { return {a.x - b.x, a.y - b.y, 0}; }
pdd operator-(const pdd &a, const pdd &b) { return sub(a, b); }
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<pdd> v;
  for (int i = 0; i < n; i++) {
    int x, y;
    cin >> x >> y;
    v.push_back({1.0 / x, 1.0 / y, i + 1});
  }
  vector<pdd> hull;
  if (n <= 3)
    hull = v;
  else {
    swap(v[0], *max_element(v.begin(), v.end(), [](const pdd &a, const pdd &b) {
           return a.y < b.y || a.y == b.y && a.x < b.x;
         }));
    sort(v.begin() + 1, v.end(), [&v](const pdd &a, const pdd &b) {
      double t = cross(a - v[0], b - v[0]);
      return t == 0 ? dot(a - v[0], a - v[0]) < dot(b - v[0], b - v[0]) : t > 0;
    });
    hull.push_back(v[0]);
    hull.push_back(v[1]);
    for (int i = 2; i < n; i++) {
      while (hull.size() > 2 &&
             cross(hull[hull.size() - 1] - hull[hull.size() - 2],
                   v[i] - hull[hull.size() - 2]) < -eps)
        hull.pop_back();
      hull.push_back(v[i]);
    }
  }
  pdd &left =
      *min_element(hull.begin(), hull.end(), [](const pdd &a, const pdd &b) {
        return a.x < b.x || a.x == b.x && a.y < b.y;
      });
  pdd &bottom =
      *min_element(hull.begin(), hull.end(), [](const pdd &a, const pdd &b) {
        return a.y < b.y || a.y == b.y && a.x < b.x;
      });
  vector<pdd> res;
  copy_if(hull.begin(), hull.end(), back_insert_iterator<vector<pdd>>(res),
          [&left, &bottom](const pdd &a) {
            return dot(left - bottom, a - bottom) > -eps &&
                   cross(left - bottom, a - bottom) > -eps;
          });
  sort(res.begin(), res.end(),
       [](const pdd &a, const pdd &b) { return a.id < b.id; });
  for (pdd &a : res) cout << a.id << ' ';
}
