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 MAXN = 2e5 + 7;
const double eps = 1e-11;
int n;
struct Pt {
  Pt() {}
  Pt(double x, double y, int id) : x(x), y(y), id(id) {}
  double x, y;
  int id;
  bool operator<(const Pt& rhs) const {
    if (make_pair(x, -y) != make_pair(rhs.x, -rhs.y))
      return make_pair(x, -y) < make_pair(rhs.x, -rhs.y);
    return id < rhs.id;
  }
} a[MAXN];
void scan() {
  cin >> n;
  for (int i = 0; i < (int)n; ++i) {
    int x, y;
    cin >> x >> y;
    a[i].x = 1.0 / x;
    a[i].y = 1.0 / y;
    a[i].id = i + 1;
  }
}
bool correct(int i1, int i2, int i3) {
  Pt p1(a[i2].x - a[i1].x, a[i2].y - a[i1].y, 0);
  Pt p2(a[i3].x - a[i2].x, a[i3].y - a[i2].y, 0);
  return p1.x * p2.y - p1.y * p2.x >= -eps;
}
void solve() {
  sort(a, a + n);
  vector<int> env;
  for (int i = 0; i < (int)n; ++i) {
    while (env.size() >= 2 && !correct(env.size() - 2, env.size() - 1, i)) {
      env.pop_back();
    }
    env.push_back(i);
  }
  vector<int> ans;
  int k = 0;
  while (k < env.size() && (k == 0 || a[env[k - 1]].y >= a[env[k]].y)) {
    ans.push_back(a[env[k]].id);
    ++k;
  }
  sort(ans.begin(), ans.end());
  for (auto x : ans) cout << x << " ";
}
int main() {
  cin.tie(NULL);
  ios_base::sync_with_stdio(0);
  scan();
  solve();
  return 0;
}
