Tavas is a cheerleader in the new sports competition named "Pashmaks".

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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;
void smain();
int main() {
  ios_base::sync_with_stdio(0);
  smain();
  return 0;
}
long long n, s[200002], r[200002];
long long pivot = 0;
inline double cw(long long p, long long a, long long b) {
  double x1 = 100.0 / s[p], y1 = 100.0 / r[p];
  double x2 = 100.0 / s[a], y2 = 100.0 / r[a];
  double x3 = 100.0 / s[b], y3 = 100.0 / r[b];
  return (x2 - x1) * (y3 - y1) - (y2 - y1) * (x3 - x1);
}
bool compare(long long a, long long b) {
  if (a == pivot) return true;
  if (b == pivot) return false;
  double t = cw(pivot, a, b);
  if (fabs(t) < 1E-9) return false;
  return t > 1E-9;
}
void smain() {
  for (long long t = 0; cin >> n; ++t) {
    for (long long i = 0; i < n; i++) cin >> s[i] >> r[i];
    for (long long i = 0; i < n; i++)
      if (s[pivot] < s[i] || (s[i] == s[pivot] && r[pivot] < r[i])) pivot = i;
    long long best = 0;
    for (long long i = 0; i < n; i++)
      if (r[best] < r[i] || (r[i] == r[best] && s[best] < s[i])) best = i;
    vector<long long> order(n);
    for (long long i = 0; i < n; i++) order[i] = i;
    sort(order.begin(), order.end(), compare);
    vector<long long> hull;
    for (long long i = 0; i < n; i++) {
      for (long long j = (long long)hull.size() - 2; j >= 0; --j) {
        if (cw(hull[j], hull.back(), order[i]) < 1E-9)
          hull.pop_back();
        else
          break;
      }
      hull.push_back(order[i]);
      if (order[i] == best) break;
    }
    sort(hull.begin(), hull.end());
    for (auto it : hull) cout << it + 1 << ' ';
    cout << '\n';
  }
}
