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 data {
  long ss, rs;
  long pos;
} d;
bool compare(const data &a, const data &b) {
  if (a.ss == b.ss)
    return a.rs < b.rs;
  else
    return a.ss < b.ss;
}
int main() {
  long n, t;
  long s, sm, r, rm, max_r = 0, max_s = 0;
  cin >> n;
  vector<data> info;
  vector<long> ser;
  for (long i = 1; i <= n; i++) {
    cin >> d.ss >> d.rs;
    d.pos = i;
    if (max_r < d.rs) max_r = d.rs, max_s = d.ss;
    if (max_r == d.rs) max_s = max(max_s, d.ss);
    info.push_back(d);
  }
  sort(info.begin(), info.end(), compare);
  ser.push_back(info[n - 1].pos);
  sm = s = info[n - 1].ss;
  rm = r = info[n - 1].rs;
  long i = n - 2;
  long double k, l;
  for (; i >= 0 && info[i].ss > max_s; i--) {
    if (info[i].rs > rm) {
      k = ((max_s - info[i].ss) * max_r * info[i].rs);
      k /= (max_s * info[i].ss * (info[i].rs - max_r));
      l = ((s - info[i].ss) * r * info[i].rs);
      l /= (s * info[i].ss * (info[i].rs - r));
      if (k > l) {
        ser.push_back(info[i].pos);
        rm = info[i].rs;
        sm = info[i].ss;
      }
    } else if (info[i].rs == rm && info[i].ss == sm) {
      ser.push_back(info[i].pos);
    }
  }
  while (info[i].ss == max_s && i >= 0) {
    if (info[i].rs == max_r) ser.push_back(info[i].pos);
    i--;
  }
  t = ser.size();
  sort(ser.begin(), ser.end());
  for (long i = 0; i < t; i++) cout << ser[i] << " ";
  return 0;
}
