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;
pair<pair<long long, long long>, long long> pts[200005];
int n;
int cnt[200005];
pair<long long, long long> intersect(pair<long long, long long> a,
                                     pair<long long, long long> b) {
  return make_pair(a.first * (b.second - a.second),
                   a.second * (b.first - a.first));
}
bool check(pair<long long, long long> a, pair<long long, long long> b) {
  return (a.first * b.second > a.second * b.first);
}
map<pair<long long, long long>, vector<int> > q;
void Readinput() {
  int x, y;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &x);
    scanf("%d", &y);
    pts[i] = pair<pair<long long, long long>, long long>(
        pair<long long, long long>(x, y), i);
    q[pts[i].first].push_back(i);
  }
}
void solve() {
  int i, j;
  vector<int> s;
  int ansx = 0, ansy = 0;
  sort(pts, pts + n);
  for (i = 0; i < n; i++) {
    while (!s.empty()) {
      int t = s.back();
      if (pts[t].first.second <= pts[i].first.second)
        s.pop_back();
      else
        break;
    }
    s.push_back(i);
  }
  vector<int> v;
  v = s;
  s.clear();
  for (int i = 0; i < v.size(); i++) {
    while (s.size() > 1 &&
           check(intersect(pts[s[s.size() - 2]].first, pts[s.back()].first),
                 intersect(pts[v[i]].first, pts[s.back()].first)))
      s.pop_back();
    s.push_back(v[i]);
  }
  v.clear();
  for (i = 0; i < s.size(); i++) {
    for (j = 0; j < q[pts[s[i]].first].size(); j++)
      v.push_back(q[pts[s[i]].first][j]);
  }
  sort(v.begin(), v.end());
  for (i = 0; i < v.size(); i++) printf("%d ", v[i] + 1);
  printf("\n");
}
int main() {
  Readinput();
  solve();
  return 0;
}
