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;
struct Vector {
  long long x, y;
  Vector(long long _x = 0, long long _y = 0) : x(_x), y(_y) {}
  inline bool operator<(const Vector& rhs) const {
    return x != rhs.x ? x > rhs.x : y > rhs.y;
  }
};
inline long long cross(Vector o, Vector a, Vector b) {
  return (a.y * b.x * o.x * o.y - a.y * o.x * b.x * b.y -
          o.y * b.x * a.x * a.y + a.x * b.x * a.y * b.y) -
         (a.x * b.y * o.x * o.y - a.x * o.y * b.x * b.y -
          o.x * b.y * a.x * a.y + a.x * b.x * a.y * b.y);
}
vector<int> ans;
map<pair<long long, long long>, vector<int>> dic;
vector<Vector> point, hull;
int main() {
  int n, s, r;
  scanf("%d", &n);
  hull.reserve(n + 5);
  point.reserve(n + 5);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &s, &r);
    dic[pair<long long, long long>(s, r)].push_back(i);
  }
  for (auto it = dic.begin(); it != dic.end(); it++)
    point.push_back(Vector(it->first.first, it->first.second));
  sort(point.begin(), point.end());
  for (int i = 0; i < (int)point.size(); i++) {
    while (hull.size() > 1 && cross(hull[(int)hull.size() - 2],
                                    hull[(int)hull.size() - 1], point[i]) < 0)
      hull.pop_back();
    hull.push_back(point[i]);
  }
  int l = -1;
  r = -1;
  for (int i = 0; i < (int)hull.size(); i++)
    if (l == -1 || hull[l].x < hull[i].x ||
        (hull[l].x == hull[i].x && hull[l].y < hull[i].y))
      l = i;
  for (int i = 0; i < (int)hull.size(); i++)
    if (r == -1 || hull[r].y < hull[i].y ||
        (hull[r].y == hull[i].y && hull[r].x < hull[i].x))
      r = i;
  for (int i = l; i <= r; i++) {
    pair<long long, long long> p =
        pair<long long, long long>(hull[i].x, hull[i].y);
    for (int& id : dic[p]) ans.push_back(id);
  }
  sort(ans.begin(), ans.end());
  for (int& id : ans) printf("%d ", id);
  puts("");
  return 0;
}
