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 int N = 200005;
struct point_comp {
  bool operator()(const complex<double> &a, const complex<double> &b) const {
    if (a.real() == b.real()) return a.imag() < b.imag();
    return a.real() < b.real();
  }
} mycomp;
inline double cross(const complex<double> &a, const complex<double> &b) {
  return imag(conj(a) * b);
}
inline double cross(const complex<double> &c, const complex<double> &a,
                    const complex<double> &b) {
  return cross(a - c, b - c);
}
vector<complex<double> > ConvexHull(vector<complex<double> > &P) {
  sort(P.begin(), P.end(), mycomp);
  int n = P.size();
  if (n < 3) return P;
  int sz = 0;
  vector<complex<double> > H(2 * n);
  for (int i = 0; i < n; i++) {
    while (sz >= 2 && cross(H[sz - 2], H[sz - 1], P[i]) <= 0) {
      --sz;
    }
    H[sz++] = P[i];
  }
  for (int i = n - 2, k = sz + 1; i >= 0; --i) {
    while (sz >= k && cross(H[sz - 2], H[sz - 1], P[i]) <= 0) {
      --sz;
    }
    H[sz++] = P[i];
  }
  H.resize(sz - 1);
  return H;
}
int main() {
  int n;
  scanf("%d", &n);
  vector<complex<double> > P(n);
  map<complex<double>, vector<int>, point_comp> ind;
  int sz = 0;
  for (int i = 0; i < n; i++) {
    double s, r;
    scanf("%lf%lf", &s, &r);
    complex<double> pt = complex<double>(1.0 / s, 1.0 / r);
    if (!ind.count(pt)) P[sz++] = pt;
    ind[pt].push_back(i + 1);
  }
  P.resize(sz);
  vector<complex<double> > H = ConvexHull(P);
  vector<int> ans;
  int cur = 0;
  do {
    vector<int> &ref = ind[H[cur]];
    ans.insert(ans.end(), ref.begin(), ref.end());
    cur++;
  } while (cur < H.size() && H[cur - 1].imag() > H[cur].imag());
  sort(ans.begin(), ans.end());
  for (int x : ans) printf("%d ", x);
  return 0;
}
