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;
int n;
set<pair<pair<int, int>, int> > st;
void init() { scanf("%d", &n); }
vector<pair<pair<int, int>, int> > v, t;
vector<int> ans;
void process() {
  st.insert(pair<pair<int, int>, int>(pair<int, int>(0, 100000), -1));
  st.insert(pair<pair<int, int>, int>(pair<int, int>(100000, 0), -1));
  int a, b;
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &a, &b);
    auto ptr =
        st.lower_bound(pair<pair<int, int>, int>(pair<int, int>(a, b), 0));
    ptr--;
    while (ptr != st.begin() && ptr->first.second < b) {
      auto tmp = ptr;
      ptr--;
      st.erase(tmp);
    }
    ptr++;
    if (ptr->first.second >= b) continue;
    st.insert(pair<pair<int, int>, int>(pair<int, int>(a, b), i + 1));
  }
  for (__typeof((st).begin()) it = (st).begin(); it != (st).end(); it++) {
    if (it->second != -1) v.push_back(*it);
  }
  for (int i = 0; i < (int)v.size(); i++) {
    pair<int, int> tmp = v[i].first;
    while (t.size() >= 2) {
      pair<int, int> t0 = t[t.size() - 1].first, t1 = t[t.size() - 2].first;
      if ((long long)t1.first * (t0.second - t1.second) *
              (t0.first * tmp.second - tmp.first * t0.second) >
          (long long)tmp.first * (tmp.second - t0.second) *
              (t1.first * t0.second - t0.first * t1.second))
        t.pop_back();
      else
        break;
    }
    t.push_back(v[i]);
  }
  for (__typeof((t).begin()) it = (t).begin(); it != (t).end(); it++)
    ans.push_back(it->second);
  sort(ans.begin(), ans.end());
  for (__typeof((ans).begin()) it = (ans).begin(); it != (ans).end(); it++)
    printf("%d ", *it);
  puts("");
}
int main() {
  init();
  process();
  return 0;
}
