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;
using db = long double;
const db eps = 1e-12;
const int N = 220000;
struct pt {
  int x, y;
  pt(int x = 0, int y = 0) : x(x), y(y) {}
};
long long vect(pt p, pt p1, pt p2) {
  return 1ll * (p1.x - p.x) * (p2.y - p.y) * p2.x * p1.y -
         1ll * (p2.x - p.x) * (p1.y - p.y) * p1.x * p2.y;
}
vector<pt> cv(vector<pt> vec) {
  if (vec.size() <= 1) return vec;
  int n = vec.size(), k = 0;
  vector<pt> ch(n + 2);
  for (int i = 0; i < n; ch[k++] = vec[i++])
    while (k >= 2 && vect(ch[k - 2], ch[k - 1], vec[i]) < 0) --k;
  while (k >= 2 && ch[k - 2].y >= ch[k - 1].y) --k;
  ch.resize(k);
  return ch;
}
map<pair<int, int>, vector<int>> M;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int s, r;
    cin >> s >> r;
    M[make_pair(s, r)].push_back(i + 1);
  }
  vector<pt> vec;
  for (auto t : M) {
    vec.emplace_back(t.first.first, t.first.second);
  }
  reverse(vec.begin(), vec.end());
  auto ret = cv(vec);
  set<int> ids;
  for (auto t : ret) {
    for (int x : M[make_pair(t.x, t.y)]) ids.insert(x);
  }
  for (int i : ids) cout << i << ' ';
}
