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 main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n;
  cin >> n;
  long long s[n], r[n];
  for (long long(i) = 0; (i) < (n); (i)++) cin >> s[i] >> r[i];
  vector<pair<long long, long long> > v, v2;
  for (long long(i) = 0; (i) < (n); (i)++) v.push_back(make_pair(s[i], r[i]));
  sort(v.begin(), v.end());
  reverse(v.begin(), v.end());
  pair<long long, long long> mx = make_pair(v[0].second, v[0].first);
  for (long long(i) = 0; (i) < (n); (i)++) {
    if (i == 0 or make_pair(v[i].second, v[i].first) > mx) {
      v2.push_back(v[i]);
      mx = make_pair(v[i].second, mx.first);
    }
  }
  stack<pair<long long, long long> > st;
  v = v2;
  for (long long(i) = 0; (i) < (v2.size()); (i)++) {
    if (st.size() < 2) {
      st.push(v2[i]);
      continue;
    }
    while (st.size() >= 2) {
      pair<long long, long long> p2 = st.top();
      st.pop();
      pair<long long, long long> p3 = st.top(), p1 = v2[i];
      if ((p2.first - p1.first) * (p2.second - p3.second) * p3.first *
              p1.second <
          (p3.first - p2.first) * (p1.second - p2.second) * p3.second *
              p1.first)
        continue;
      st.push(p2);
      break;
    }
    st.push(v2[i]);
  }
  map<pair<long long, long long>, long long> dic;
  while (st.size()) {
    pair<long long, long long> tmp = st.top();
    st.pop();
    dic[tmp] = 1;
  }
  for (long long(i) = 0; (i) < (n); (i)++) {
    if (dic[make_pair(s[i], r[i])] > 0) cout << i + 1 << " ";
  }
  cout << "\n";
  return 0;
}
