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;
stack<pair<pair<long long, long long>, int> > st;
pair<pair<long long, long long>, int> v[200010], a[200010];
bool never[200010];
int rez[200010];
inline long long gcd(long long a, long long b) {
  for (; b;) {
    a %= b;
    swap(a, b);
  }
  return a;
}
inline pair<long long, long long> obt(pair<long long, long long> a,
                                      pair<long long, long long> b) {
  long long aa = (a.first - b.first) * a.second * b.second;
  long long bb = a.first * b.first * (b.second - a.second);
  long long cc = 1LL;
  return {aa / cc, bb / cc};
}
inline bool solve(pair<long long, long long> in, pair<long long, long long> a,
                  pair<long long, long long> b) {
  pair<long long, long long> alt = obt(a, b);
  return (alt.first * in.second <= alt.second * in.first);
}
inline bool cmp(pair<pair<long long, long long>, int> a,
                pair<pair<long long, long long>, int> b) {
  if (a.first.second == b.first.second) return (a.first.first < b.first.first);
  return (a.first.second < b.first.second);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i)
    scanf("%I64d %I64d", &v[i].first.first, &v[i].first.second),
        v[i].second = i;
  sort(v + 1, v + n + 1, cmp);
  v[0].first.first = 2000000000LL;
  int m = 0;
  for (int i = 1; i <= n; ++i) {
    if (v[i].first.first == v[i - 1].first.first) {
      if (v[i].first == v[i - 1].first)
        never[v[i].second] = never[v[i - 1].second] = true;
      continue;
    }
    a[++m] = v[i];
  }
  st.push(make_pair(make_pair(-20000LL, 1LL), 1));
  for (int i = 2; i <= m; ++i) {
    for (; st.size() > 1;) {
      if (solve(st.top().first, a[st.top().second].first, a[i].first))
        st.pop();
      else
        break;
    }
    st.push(make_pair(obt(a[st.top().second].first, a[i].first), i));
  }
  m = 0;
  pair<pair<long long, long long>, int> lst;
  lst.first.first = lst.first.second = 1LL;
  for (; st.size() > 0; lst = st.top(), st.pop())
    if (!never[a[st.top().second].second] &&
        lst.first.first * lst.first.second > 0)
      rez[++m] = a[st.top().second].second;
  sort(rez + 1, rez + m + 1);
  for (int i = 1; i <= m; ++i) printf("%d ", rez[i]);
  if (!m) printf("-1");
  printf("\n");
  return 0;
}
