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 long long mod = 1e9 + 7;
const int mn = 3e5 + 10;
struct Line {
  long long a, b;
  int ind;
};
__int128 one = 1;
bool ded(Line a, Line b, Line c) {
  return one * a.b * c.a * (b.b - c.b) * (b.a - a.a) <
         one * a.a * c.b * (a.b - b.b) * (c.a - b.a);
}
deque<Line> cvx;
int p[mn];
int f(int x) { return x == p[x] ? x : (p[x] = f(p[x])); }
bool gud[mn];
int main() {
  cin.tie(0), cin.sync_with_stdio(0);
  iota(p, p + mn, 0);
  int n;
  scanf("%d", &n);
  vector<Line> v;
  for (int i = 1; i <= n; i++) {
    long long a, b;
    scanf("%lld%lld", &a, &b);
    v.push_back({a, b, i});
  }
  sort(v.begin(), v.end(), [](Line a, Line b) {
    if (a.a != b.a) return a.a < b.a;
    return a.b > b.b;
  });
  for (Line l : v) {
    if (cvx.size() && cvx.back().a == l.a) {
      if (l.b == cvx.back().b) p[l.ind] = f(cvx.back().ind);
      continue;
    }
    while (cvx.size() >= 2 && ded(cvx[cvx.size() - 2], cvx[cvx.size() - 1], l))
      cvx.pop_back();
    cvx.push_back(l);
  }
  while (cvx[0].b - cvx[1].b <= 0) cvx.pop_front();
  for (Line l : cvx) gud[l.ind] = 1;
  for (int i = 1; i <= n; i++)
    if (gud[f(i)]) printf("%d ", i);
}
