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("Ofast,unroll-loops,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const int maxn = 2e5 + 10;
vector<long double> m, b, _m, _b;
bool bad(int l1, int l2, int l3) {
  return ((long double)10000 / b[l3] - (long double)10000 / b[l1]) *
             ((long double)10000 / m[l1] - (long double)10000 / m[l2]) <
         ((long double)10000 / b[l2] - (long double)10000 / b[l1]) *
             ((long double)10000 / m[l1] - (long double)10000 / m[l3]);
}
pair<long double, long double> p[maxn];
vector<pair<long double, long double> > v, x;
int main(int argc, char const *argv[]) {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    p[i].first = a;
    p[i].second = b;
    v.push_back(p[i]);
  }
  sort(v.begin(), v.end());
  for (int i = 0; i < v.size(); i++) {
    x.push_back(v[i]);
    while (x.size() >= 2 && x[x.size() - 2].second <= x.back().second)
      x.erase(x.end() - 2);
  }
  for (auto it : x) {
    m.push_back(it.first);
    b.push_back(it.second);
    int s = m.size();
    while (s >= 3 && bad(s - 3, s - 2, s - 1)) {
      m.erase(m.end() - 2);
      b.erase(b.end() - 2);
      s--;
    }
  }
  set<pair<long double, long double> > st;
  for (int i = 0; i < m.size(); i++) {
    st.insert({m[i], b[i]});
  }
  for (int i = 0; i < n; i++) {
    if (st.find(p[i]) != st.end()) printf("%d ", i + 1);
  }
}
