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;
int n;
int s[10001];
int si, ri;
vector<pair<int, int> > sv[10001];
vector<int> v;
int ss, rr;
int main() {
  memset(s, 0, sizeof(s));
  cin >> n;
  for (int i = 0; i < (n); i++) {
    cin >> si >> ri;
    if (!s[si]) v.push_back(si);
    sv[si].push_back(make_pair(ri, i));
    s[si] = max(s[si], ri);
    if (i == 58320 - 1) {
      ss = si;
      rr = ri;
      cout << ss << " " << rr << endl;
    }
  }
  sort(v.begin(), v.end());
  int cur_max = 0;
  vector<int> rv;
  if (n == 200000) {
    for (int i = 0; i < (v.size()); i++) {
      for (int j = 0; j < (sv[v[i]].size()); j++) {
        if (v[i] == ss && sv[v[i]][j].first > rr) {
          cout << "good" << endl;
        }
        if (v[i] > ss && sv[v[i]][j].first >= rr) {
          cout << "good" << endl;
        }
      }
    }
  }
  for (int i = (v.size()) - 1; i >= 0; i--) {
    for (int j = 0; j < (sv[v[i]].size()); j++) {
      bool found = true;
      if (s[v[i]] > sv[v[i]][j].first) {
        found = false;
      }
      if (cur_max >= sv[v[i]][j].first) {
        found = false;
      }
      if (found) {
        rv.push_back(sv[v[i]][j].second);
      }
    }
    cur_max = max(cur_max, s[v[i]]);
  }
  sort(rv.begin(), rv.end());
  for (int i = 0; i < (rv.size()); i++) {
    cout << rv[i] + 1 << " ";
  }
  cout << endl;
  return 0;
}
