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 long long INF = 1e9 + 7;
const int N = 1e6 + 10;
struct Node {
  int x, y, pos;
  void init(int o) {
    cin >> x >> y;
    pos = o;
  }
  bool operator<(const Node &node) const {
    if (x != node.x) return x > node.x;
    return y > node.y;
  }
};
vector<Node> v;
bool check(int i, int j, int k) {
  int s1 = v[i].x, t1 = v[i].y;
  int s2 = v[j].x, t2 = v[j].y;
  int s3 = v[k].x, t3 = v[k].y;
  if (s1 == s2 && t1 == t2) return true;
  if (s2 == s3 && t2 == t3) return true;
  return 1LL * (t3 - t2) * s3 * (s1 - s2) * t1 <=
         1LL * (t2 - t1) * s1 * (s2 - s3) * t3;
}
int main() {
  int n;
  cin >> n;
  v.resize(n);
  for (int i = 1; i <= n; i++) v[i - 1].init(i);
  sort(v.begin(), v.end());
  vector<int> ans;
  int x = 0, y = 0;
  for (int i = 0; i < v.size(); i++) {
    auto &node = v[i];
    if (x == node.x && y == node.y) {
      continue;
    }
    if (x >= node.x && y >= node.y) {
      continue;
    }
    while (ans.size() >= 2 &&
           !check(ans[ans.size() - 2], ans[ans.size() - 1], i))
      ans.pop_back();
    ans.push_back(i);
    x = node.x;
    y = node.y;
  }
  map<pair<int, int>, int> mp;
  for (auto &x : ans) {
    mp[{v[x].x, v[x].y}] = 1;
  }
  ans.clear();
  for (int i = 0; i < n; i++) {
    if (mp.count({v[i].x, v[i].y})) ans.push_back(v[i].pos);
  }
  sort(ans.begin(), ans.end());
  for (auto &x : ans) {
    printf("%d ", x);
  }
  return 0;
}
