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;
const int N = 200100;
multimap<pair<int, int>, int> m;
pair<double, double> p[N], s[N];
int ans[N];
int dcmp(double x) {
  if (x > 10e-8) return 1;
  if (x < -10e-8) return -1;
  return 0;
}
pair<double, double> operator+(pair<double, double> a, pair<double, double> b) {
  return pair<double, double>(a.first + b.first, a.second + b.second);
}
pair<double, double> operator-(pair<double, double> a, pair<double, double> b) {
  return pair<double, double>(a.first - b.first, a.second - b.second);
}
double cross(pair<double, double> a, pair<double, double> b) {
  return a.first * b.second - a.second * b.first;
}
double cross(pair<double, double> a, pair<double, double> b,
             pair<double, double> c) {
  return cross(b - a, c - a);
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int t, v;
    scanf("%d%d", &t, &v);
    p[i].first = 10e4 / t;
    p[i].second = 10e4 / v;
    m.insert(pair<pair<int, int>, int>(pair<int, int>(t, v), i));
  }
  sort(p + 1, p + n + 1);
  int cnt = 0;
  for (int i = 1; i <= n; i++)
    if (p[i] != p[i - 1]) {
      while (cnt >= 2 && dcmp(cross(s[cnt - 1], s[cnt], p[i])) < 0) cnt--;
      s[++cnt] = p[i];
    }
  while (cnt >= 2 && dcmp(s[cnt].second - s[cnt - 1].second) >= 0) cnt--;
  int num = 0;
  for (int i = 1; i <= cnt; i++) {
    pair<int, int> cur((int)(10e4 / s[i].first + 0.5),
                       (int)(10e4 / s[i].second + 0.5));
    multimap<pair<int, int>, int>::iterator l = m.lower_bound(cur);
    multimap<pair<int, int>, int>::iterator r = m.upper_bound(cur);
    for (multimap<pair<int, int>, int>::iterator j = l; j != r; j++)
      ans[num++] = j->second;
  }
  sort(ans, ans + num);
  for (int i = 0; i <= num - 1; i++) printf("%d ", ans[i]);
  return 0;
}
