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 <typename A, typename B>
inline void chmin(A &a, B b) {
  if (a > b) a = b;
}
template <typename A, typename B>
inline void chmax(A &a, B b) {
  if (a < b) a = b;
}
long long N;
long long S[222222], R[222222];
bool ushi(pair<long long, long long> a, pair<long long, long long> b,
          pair<long long, long long> c) {
  return (a.first * c.first - b.first * c.first) *
             (a.second * b.second - b.second * c.second) >=
         (a.second * c.second - b.second * c.second) *
             (a.first * b.first - b.first * c.first);
}
signed main() {
  scanf("%lld", &N);
  for (long long i = 0; i < (N); i++) scanf("%lld%lld", &S[i], &R[i]);
  vector<pair<long long, long long> > v;
  for (long long i = 0; i < (N); i++)
    v.push_back(pair<long long, long long>(S[i], R[i]));
  sort((v).begin(), (v).end());
  reverse((v).begin(), (v).end());
  vector<pair<long long, long long> > u;
  for (auto p : v) {
    if (u.size() && u.back().second >= p.second) continue;
    u.push_back(p);
  }
  reverse((u).begin(), (u).end());
  vector<pair<long long, long long> > st;
  for (long long i = (long long)u.size() - 1; i >= 0; i--) {
    while (st.size() >= 2 && !ushi(st[st.size() - 2], st.back(), u[i]))
      st.pop_back();
    st.push_back(u[i]);
  }
  sort((st).begin(), (st).end());
  vector<long long> ans;
  for (long long i = 0; i < (N); i++) {
    if (binary_search((st).begin(), (st).end(),
                      pair<long long, long long>(S[i], R[i])))
      ans.push_back(i);
  }
  for (long long i = 0; i < (ans.size()); i++) {
    if (i) printf(" ");
    printf("%lld", ans[i] + 1);
  }
  puts("");
  return 0;
}
