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 pt {
  long double x, y;
  pt(long double x, long double y) : x(x), y(y) {}
  pt() {}
  long double norm2() { return *this * *this; }
  long double norm() { return sqrt(norm2()); }
  bool operator==(pt p) { return abs(x - p.x) < 1e-12 && abs(y - p.y) < 1e-12; }
  pt operator+(pt p) { return pt(x + p.x, y + p.y); }
  pt operator-(pt p) { return pt(x - p.x, y - p.y); }
  pt operator*(long double t) { return pt(x * t, y * t); }
  pt operator/(long double t) { return pt(x / t, y / t); }
  long double operator*(pt p) { return x * p.x + y * p.y; }
  long double angle(pt p) { return acos(*this * p / (norm() * p.norm())); }
  pt unit() { return *this / norm(); }
  long double operator%(pt p) { return x * p.y - y * p.x; }
  bool operator<(pt p) const {
    return x < p.x || (abs(x - p.x) < 1e-12 && y < p.y);
  }
  bool left(pt p, pt q) { return (q - p) % (*this - p) >= 0; }
  pt rot(pt r) { return pt(*this % r, *this * r); }
  pt rot(long double a) { return rot(pt(sin(a), cos(a))); }
};
pt ccw90(1, 0);
pt cw90(-1, 0);
vector<pt> chull(vector<pt> p) {
  if (((int)p.size()) < 3) return p;
  vector<pt> r;
  sort(p.begin(), p.end());
  for (int i = 0, ThxDem = p.size(); i < ThxDem; ++i) {
    while (r.size() >= 2 && r.back().left(r[r.size() - 2], p[i])) r.pop_back();
    r.push_back(p[i]);
  }
  r.pop_back();
  int k = r.size();
  for (int i = p.size() - 1; i >= 0; --i) {
    while (r.size() >= k + 2 && r.back().left(r[r.size() - 2], p[i]))
      r.pop_back();
    r.push_back(p[i]);
  }
  r.pop_back();
  return r;
}
int n, m;
vector<pt> pingo;
map<pair<int, int>, vector<int> > ereccion;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int xi, yi;
    scanf("%d %d", &xi, &yi);
    pingo.push_back(pt(1. / xi, 1. / yi));
    ereccion[make_pair(xi, yi)].push_back(i + 1);
  }
  vector<pt> concha = chull(pingo);
  vector<pt> ura, pito;
  for (auto p : concha) {
    ura.push_back(pt(p.x, p.y));
    pito.push_back(pt(p.y, p.x));
  }
  sort(ura.begin(), ura.end());
  pt leftmost = pt(ura[0].x, ura[0].y);
  sort(pito.begin(), pito.end());
  pt downmost = pt(pito[0].y, pito[0].x);
  set<int> ans;
  for (auto p : concha) {
    if (p.x - 1e-12 < downmost.x && p.y - 1e-12 < leftmost.y &&
        downmost.left(leftmost, p)) {
      int xx = (1. / p.x + 1e-3);
      int yy = (1. / p.y + 1e-3);
      assert(ereccion[make_pair(xx, yy)].size());
      for (auto d : ereccion[make_pair(xx, yy)]) ans.insert(d);
    }
  }
  for (auto d : ans) {
    printf("%d ", d);
  }
  puts("");
  return 0;
}
