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;
template <class T>
T abs(T x) {
  return x > 0 ? x : -x;
}
template <class T>
T gcd(T a, T b) {
  return a ? gcd(b % a, a) : b;
}
template <class T>
T sqr(T a) {
  return a * a;
}
template <class T>
T sgn(T a) {
  return a > 0 ? 1 : (a < 0 ? -1 : 0);
}
int n;
int m;
vector<pair<pair<int, int>, int>> a;
long long vect(pair<int, int> a, pair<int, int> b, pair<int, int> c) {
  return ((long long)b.first - a.first) * ((long long)c.second - a.second) *
             (long long)b.second * c.first -
         ((long long)b.second - a.second) * ((long long)c.first - a.first) *
             (long long)b.first * c.second;
}
long long vect(int i, int j, int k) {
  return vect(a[i].first, a[j].first, a[k].first);
}
int main() {
  scanf("%d", &n);
  pair<pair<int, int>, int> right = make_pair(make_pair(-1, -1), -1),
                            down = make_pair(make_pair(-1, -1), -1);
  map<pair<int, int>, vector<int>> b;
  for (int i = 0; i < n; i++) {
    int r, s;
    scanf("%d%d", &r, &s);
    if (b.count(make_pair(r, s)) == 0) {
      a.push_back(make_pair(make_pair(r, s), i));
      right = max(make_pair(make_pair(r, s), i), right);
      down = max(make_pair(make_pair(s, r), i), down);
    }
    b[make_pair(r, s)].push_back(i);
  }
  sort((a).begin(), (a).end(),
       [](const pair<pair<int, int>, int>& a,
          const pair<pair<int, int>, int>& b) {
         int v =
             b.first.first * a.first.second - a.first.first * b.first.second;
         return v > 0 ||
                (v == 0 &&
                 1.0 / sqr(a.first.first) + 1.0 / sqr(a.first.second) >
                     1.0 / sqr(b.first.first) + 1.0 / sqr(b.first.second));
       });
  int i = 0;
  for (; i < n && a[i].second != down.second;) i++;
  vector<int> st;
  for (; i < n; i++) {
    while (((int)(st).size()) > 1 &&
           (vect(*-- --st.end(), *--st.end(), i) > 0 ||
            a[*--st.end()].first == a[i].first))
      st.pop_back();
    st.push_back(i);
    if (a[i].second == right.second) break;
  }
  vector<int> ans;
  for (auto& cur : st) {
    for (auto q : b[a[cur].first]) {
      ans.push_back(q + 1);
    }
  }
  sort((ans).begin(), (ans).end());
  for (auto cur : ans) printf("%d ", cur);
  return 0;
}
