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 val[100005], n;
vector<int> id[100005];
vector<pair<int, int> > vec;
bool check(pair<int, int> x, pair<int, int> y, pair<int, int> z) {
  long long a = (long long)x.first, b = (long long)x.second;
  long long c = (long long)y.first, d = (long long)y.second;
  long long e = (long long)z.first, f = (long long)z.second;
  return (a * b - b * c) * (e * f - d * e) > (a * d - a * b) * (c * f - e * f);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    if (val[a] < b) {
      val[a] = b;
      id[a].clear();
      id[a].push_back(i);
    } else if (val[a] == b)
      id[a].push_back(i);
  }
  for (int i = 10000; i >= 1; i--) {
    if (!id[i].size()) continue;
    vec.push_back(make_pair(i, val[i]));
  }
  deque<pair<int, int> > deq;
  for (int i = 0; i < vec.size(); i++) {
    if (!deq.empty() && vec[i].second <= deq.back().second) continue;
    while (deq.size() >= 2 &&
           check(deq[deq.size() - 2], deq[deq.size() - 1], vec[i]))
      deq.pop_back();
    deq.push_back(vec[i]);
  }
  vector<int> res;
  for (int i = 0; i < deq.size(); i++) {
    for (int j = 0; j < id[deq[i].first].size(); j++)
      res.push_back(id[deq[i].first][j]);
  }
  sort(res.begin(), res.end());
  for (int i = 0; i < res.size(); i++)
    printf("%d%c", res[i], i == res.size() - 1 ? '\n' : ' ');
}
