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>
#pragma GCC optimize("O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
using namespace std;
struct R {
  R() : first(0), second(0) {}
  R(long long _x, long long _y) : first(_x), second(_y) {}
  long long first, second;
};
inline R operator+(const R& a, const R& b) {
  return R({a.first * b.second + b.first * a.second, a.second * b.second});
}
inline R operator-(const R& a, const R& b) {
  auto ans = R({a.first * b.second - b.first * a.second, a.second * b.second});
  if (ans.second < 0) {
    ans.first = -ans.first;
    ans.second = -ans.second;
  }
  return ans;
}
inline R operator*(const R& a, const R& b) {
  return R({a.first * b.first, a.second * b.second});
}
inline bool operator<(const R& a, const R& b) { return (a - b).first < 0; }
inline bool operator==(const R& a, const R& b) {
  return a.first == b.first && a.second == b.second;
}
template <class T>
pair<T, T> operator-(const pair<T, T>& a, const pair<T, T>& b) {
  return make_pair(a.first - b.first, a.second - b.second);
}
template <class T>
T vec(const pair<T, T>& a, const pair<T, T>& b) {
  return a.first * b.second - b.first * a.second;
}
template <class T>
T vec(const pair<pair<T, T>, int>& a, const pair<pair<T, T>, int>& b) {
  return vec(a.first, b.first);
}
int a, b, c, d, n, m, k;
pair<pair<int, int>, int> mas[200002];
int main() {
  scanf("%d", &n);
  for (int _n((n)-1), i(0); i <= _n; i++) {
    scanf("%d%d", &mas[i].first.first, &mas[i].first.second);
    mas[i].second = i;
  }
  vector<pair<R, R>> pnt;
  for (int _n((n)-1), i(0); i <= _n; i++) {
    R first({1LL, (long long)mas[i].first.first});
    R second({1LL, (long long)mas[i].first.second});
    auto t = make_pair(first, second);
    pnt.push_back(t);
  }
  sort((pnt).begin(), (pnt).end());
  vector<pair<R, R>> hull;
  for (int _n((n)-1), i(0); i <= _n; i++) {
    if (i && pnt[i - 1].first == pnt[i].first) continue;
    auto t = pnt[i];
    while (hull.size() > 1 && vec(hull.back() - hull[(int)((hull).size()) - 2],
                                  t - hull[(int)((hull).size()) - 2])
                                      .first < 0)
      hull.pop_back();
    hull.push_back(t);
  }
  sort((hull).begin(), (hull).end());
  vector<int> res;
  for (int _n((n)-1), i(0); i <= _n; i++) {
    R first({1LL, (long long)mas[i].first.first});
    R second({1LL, (long long)mas[i].first.second});
    auto t = make_pair(first, second);
    if (binary_search((hull).begin(), (hull).end(), t)) {
      res.push_back(mas[i].second);
    }
  }
  sort((res).begin(), (res).end());
  for (int _n(((int)((res).size())) - 1), i(0); i <= _n; i++) {
    if (i) printf(" ");
    printf("%d", res[i] + 1);
  }
  printf("\n");
}
