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 ll = long long;
using ull = unsigned long long;
using ld = long double;
const ld M_PI = acos(-1.0);
const ld eps = 1e-9;
const int INF = 2000000000;
const ll LINF = 1ll * INF * INF;
const ll MOD = 1000000007;
struct Point {
  ll s, r;
  int ix;
  Point(){};
  Point(ll _s, ll _r) : s(_s), r(_r), ix(-1){};
  bool operator<(const Point& x) const {
    if (s == x.s) return r > x.r;
    return s > x.s;
  }
};
bool good(Point a, Point b, Point c) {
  return ((b.s - c.s) * (a.r - b.r) * a.s * c.r -
              (b.r - c.r) * (a.s - b.s) * a.r * c.s <=
          0);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<Point> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i].s >> a[i].r;
    a[i].ix = i;
  }
  sort((a).begin(), (a).end());
  vector<Point> q;
  for (int i = 0; i < n; ++i) {
    int sz = q.size();
    while (sz >= 2 && !good(q[sz - 2], q[sz - 1], a[i])) {
      q.pop_back();
      --sz;
    }
    q.emplace_back(a[i]);
  }
  int sz = q.size();
  while (sz >= 2 && q[sz - 1].r <= q[sz - 2].r) {
    q.pop_back();
    --sz;
  }
  vector<int> ans;
  for (Point p : q) ans.emplace_back(p.ix);
  sort((ans).begin(), (ans).end());
  for (int i : ans) cout << 1 + i << ' ';
}
