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;
int n;
pair<pair<long long, long long>, int> comp[200002];
bool cross(pair<long long, long long> a, pair<long long, long long> b,
           pair<long long, long long> c) {
  return (a.second * b.second - a.second * c.second) *
                 (b.first * c.first - a.first * c.first) <
             (a.second * c.second - b.second * c.second) *
                 (a.first * c.first - a.first * b.first) ||
         c.second >= b.second;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%I64d %I64d", &comp[i].first.first, &comp[i].first.second);
    comp[i].second = i;
  }
  comp[n] = make_pair(make_pair(0, 0), n);
  sort(comp, comp + n);
  vector<int> hull;
  for (int i = 0; i < n; i++) {
    while (hull.size() >= 2 && cross(comp[hull[hull.size() - 2]].first,
                                     comp[hull.back()].first, comp[i].first))
      hull.pop_back();
    hull.push_back(i);
  }
  vector<pair<long long, long long> > ans;
  while (hull.size() >= 2) {
    pair<long long, long long> bef = comp[hull[hull.size() - 2]].first;
    pair<long long, long long> now = comp[hull.back()].first;
    if (bef.second <= now.second && bef != now) break;
    ans.push_back(comp[hull.back()].first);
    hull.pop_back();
  }
  ans.push_back(comp[hull.back()].first);
  set<pair<long long, long long> > super;
  while (!ans.empty()) {
    super.insert(ans.back());
    ans.pop_back();
  }
  vector<int> tans;
  for (int i = 0; i < n; i++) {
    if (super.find(comp[i].first) != super.end()) {
      tans.push_back(comp[i].second);
    }
  }
  sort(tans.begin(), tans.end());
  for (int i = 0; i < tans.size(); i++) printf("%d ", tans[i] + 1);
  printf("\n");
  return 0;
}
