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;
double eps = 1e9;
int a[10010] = {0};
bool b[10010] = {0};
pair<int, int> q[200100];
class Line {
 public:
  int minv, cinv;
};
bool operator<(Line x, Line y) { return x.minv < y.minv; }
vector<Line> v;
bool invalid(Line a, Line b, Line c) {
  double x1;
  double p = 1.0 / b.cinv - 1.0 / a.cinv;
  double q = 1.0 / a.minv - 1.0 / b.minv;
  x1 = p / q;
  p = 1.0 / c.cinv - 1.0 / b.cinv;
  q = 1.0 / b.minv - 1.0 / c.minv;
  double x2 = p / q;
  if (x2 < (x1 - eps)) return 1;
  return 0;
}
int main() {
  int n;
  cin >> n;
  int x, y;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &x, &y);
    a[x] = max(a[x], y);
    q[i] = pair<int, int>(x, y);
  }
  int maxi = 0;
  for (int i = 1; i <= 10000; i++) {
    if (a[i] > 0) {
      maxi = max(maxi, a[i]);
      Line l;
      l.minv = i;
      l.cinv = a[i];
      v.push_back(l);
    }
  }
  sort(v.begin(), v.end());
  stack<Line> s;
  for (int i = 0; i < v.size(); i++) {
    Line l = v[i];
    s.push(l);
    while (1) {
      if (s.size() <= 2) break;
      Line cc = s.top();
      s.pop();
      Line bb = s.top();
      s.pop();
      Line aa = s.top();
      s.pop();
      if (invalid(aa, bb, cc)) {
        s.push(aa);
        s.push(cc);
      } else {
        s.push(aa);
        s.push(bb);
        s.push(cc);
        break;
      }
    }
  }
  vector<int> u;
  while (!s.empty()) {
    Line l = s.top();
    s.pop();
    u.push_back(l.minv);
  }
  reverse(u.begin(), u.end());
  bool ok = 0;
  for (int i = 0; i < u.size(); i++) {
    if (ok || a[u[i]] == maxi) ok = 1;
    if (ok) b[u[i]] = 1;
  }
  for (int i = 1; i <= n; i++) {
    if (b[q[i].first] && a[q[i].first] == q[i].second) cout << i << " ";
  }
  cout << endl;
}
