Tavas is a cheerleader in the new sports competition named "Pashmaks".

<image>

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 maxn = 200050;
const double eps = 1e-8;
inline int dcmp(double first) {
  if (first > eps)
    return 1;
  else if (first < -eps)
    return -1;
  return 0;
}
struct Point {
  int id, first, second;
  bool operator<(const Point &rhs) const {
    return make_pair(-first, second) < make_pair(-rhs.first, rhs.second);
  }
  bool operator==(const Point &rhs) const {
    return make_pair(first, second) == make_pair(rhs.first, rhs.second);
  }
  void scan(int i) { id = i, scanf("%d%d", &first, &second); }
};
int n, ans[maxn], anss = 0;
Point p[maxn];
int st[maxn], top = 0;
pair<double, double> operator-(Point A, Point B) {
  return make_pair(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;
}
int main() {
  int i, j;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) p[i].scan(i);
  sort(p + 1, p + 1 + n);
  for (i = 1; i <= n; i = j) {
    while (top > 1 && 1LL * p[i].first * p[st[top - 1]].second *
                              (p[st[top - 1]].first - p[st[top - 2]].first) *
                              (p[i].second - p[st[top - 2]].second) <
                          1LL * p[i].second * p[st[top - 1]].first *
                              (p[st[top - 1]].second - p[st[top - 2]].second) *
                              (p[i].first - p[st[top - 2]].first))
      top--;
    st[top++] = i;
    for (j = i + 1; j <= n && p[j] == p[i]; j++)
      ;
  }
  for (i = 1, j = 0; i <= n && j < top; j++) {
    if (j && p[st[j]].second < p[st[j - 1]].second) break;
    while (st[j] != i) i++;
    while (i <= n && p[st[j]] == p[i]) ans[anss++] = p[i++].id;
  }
  sort(ans, ans + anss);
  for (i = 0; i < anss; i++) printf("%d%c", ans[i], i < anss - 1 ? ' ' : '\n');
  return 0;
}
