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-30;
int n;
struct Nodo {
  double x, y;
  int idx;
  Nodo() {}
  Nodo(const double &_x, const double &_y, const int &_idx) {
    x = _x;
    y = _y;
    idx = _idx;
  }
};
bool operator<(const Nodo &a, const Nodo &b) {
  if (fabs(a.x - b.x) < eps) return a.y < b.y;
  return a.x < b.x;
}
Nodo arr[200005];
vector<Nodo> res;
vector<int> sol;
double cross(const Nodo p0, const Nodo p1, const Nodo p2) {
  return (p1.x - p0.x) * (p2.y - p0.y) - (p2.x - p0.x) * (p1.y - p0.y);
}
int main() {
  scanf("%d", &n);
  int s, r;
  for (int i = 0; i < n; ++i) {
    scanf("%d%d", &s, &r);
    arr[i] = Nodo(1. / s, 1. / r, i + 1);
  }
  if (n == 1) {
    puts("1");
    return 0;
  }
  sort(arr, arr + n);
  res.push_back(arr[0]);
  res.push_back(arr[1]);
  int sz = 0;
  double miny = min(arr[0].y, arr[1].y);
  for (int i = 2; i < n; ++i) {
    miny = min(miny, arr[i].y);
    if (cross(res[sz], res[sz + 1], arr[i]) > -eps) {
      res.push_back(arr[i]);
      ++sz;
    } else {
      res.pop_back();
      if (sz == 0)
        res.push_back(arr[i]);
      else {
        --sz;
        --i;
      }
    }
  }
  for (auto x : res) {
    sol.push_back(x.idx);
    if (fabs(miny - x.y) < eps) break;
  }
  sort(sol.begin(), sol.end());
  for (auto it : sol) {
    printf("%d ", it);
  }
  puts("");
  return 0;
}
