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;
const int MAXN = (1 << 20);
const double eps = 1e-22;
int n;
pair<int, pair<int, int> > a[MAXN];
void read() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i].first >> a[i].second.first;
    a[i].second.second = i + 1;
  }
}
map<pair<int, int>, bool> has;
double slope(pair<int, pair<int, int> > a, pair<int, pair<int, int> > b) {
  return (double)(a.second.first - b.second.first) /
         (double)(a.first - b.first);
}
void solve() {
  vector<pair<int, pair<int, int> > > li;
  sort(a, a + n);
  int mx = 0;
  for (int i = n - 1; i >= 0; i--) {
    if (mx < a[i].second.first) li.push_back(a[i]);
    mx = max(mx, a[i].second.first);
  }
  sort(li.begin(), li.end());
  vector<pair<int, pair<int, int> > > st;
  st.push_back(li.back());
  for (int i = li.size() - 2; i >= 0; i--) {
    while (st.size() >= 2 &&
           slope(st[st.size() - 2], li[i]) -
                   slope(st[st.size() - 2], st[st.size() - 1]) >
               eps)
      st.pop_back();
    st.push_back(li[i]);
  }
  for (int pos = 0; pos < st.size(); pos++)
    has[make_pair(st[pos].first, st[pos].second.first)] = 1;
  vector<int> ans;
  for (int i = 0; i < n; i++)
    if (has[make_pair(a[i].first, a[i].second.first)])
      ans.push_back(a[i].second.second);
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  read();
  solve();
  return 0;
}
