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;
vector<pair<int, int> > p;
bool cmp(int g, int h) {
  return (p[g].first > p[h].first) ||
         (p[g].first == p[h].first && p[g].second > p[h].second);
}
int main() {
  int n;
  scanf("%d", &n);
  vector<int> num;
  for (int i = 0; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    p.push_back(make_pair(x, -y));
    num.push_back(i);
  }
  sort(num.begin(), num.end(), cmp);
  vector<int> c;
  for (int i = 0; i < n; i++) {
    while (c.size() && p[num[i]].first == p[c.back()].first) c.pop_back();
    while (c.size() > 1) {
      int g = c[c.size() - 2];
      int h = c[c.size() - 1];
      int j = num[i];
      long long k1 = p[h].second * p[j].first;
      long long k2 = p[h].first * p[j].second;
      if (k1 * (p[h].first - p[g].first) * (p[j].second - p[g].second) -
              k2 * (p[h].second - p[g].second) * (p[j].first - p[g].first) <=
          0)
        c.pop_back();
      else
        break;
    }
    c.push_back(num[i]);
  }
  while (c.size() > 1 && p[c.back()].second > p[c[c.size() - 2]].second)
    c.pop_back();
  sort(c.begin(), c.end());
  for (int i = 0; i < c.size(); i++)
    printf("%d%c", c[i] + 1, " \n"[i == c.size() - 1]);
  return 0;
}
