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;
const long long MAXN = 2e5 + 7;
long long n;
pair<long double, long double> a[MAXN];
vector<pair<long double, long double> > ans;
pair<long double, long double> vect(pair<long double, long double> a,
                                    pair<long double, long double> b) {
  return {b.first - a.first, b.second - a.second};
}
long double cp(pair<long double, long double> a,
               pair<long double, long double> b) {
  return a.first * b.second - a.second * b.first;
}
bool comp(long long i, long long j) { return a[i] < a[j]; }
void add(pair<long double, long double> p) {
  while (2 <= ans.size()) {
    auto prpr = ans[(long long)ans.size() - 2];
    auto pr = ans[(long long)ans.size() - 1];
    if (cp(vect(prpr, p), vect(prpr, pr)) < 0) {
      ans.pop_back();
    } else {
      break;
    }
  }
  while (ans.size() && ans.back() != p && ans.back().first <= p.first &&
         ans.back().second <= p.second)
    ans.pop_back();
  ans.push_back(p);
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (long long i = 0; i < n; ++i) {
    long long first, second;
    cin >> first >> second;
    a[i].first = 1 - 1.0 / first;
    a[i].second = 1 - 1.0 / second;
  }
  set<pair<long double, long double> > ms;
  for (long long i = 0; i < n; ++i) ms.insert(a[i]);
  for (auto e : ms) add(e);
  ms.clear();
  for (auto e : ans) ms.insert(e);
  for (long long i = 0; i < n; ++i) {
    if (ms.find(a[i]) != ms.end()) cout << i + 1 << ' ';
  }
  cout << '\n';
  return 0;
}
