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 
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200100;
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> > chull(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;
  cin >> N;
  vector<complex<double> > P(N);
  map<complex<double>, vector<int>, point_comp> ind;
  int sz = 0;
  for (int i = 0; i < N; ++i) {
    int S, R;
    cin >> 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 = chull(P);
  int cur = 0;
  vector<int> R;
  do {
    vector<int> &ref = ind[H[cur]];
    R.insert(R.end(), ref.begin(), ref.end());
    ++cur;
  } while (cur < H.size() && H[cur - 1].imag() > H[cur].imag());
  sort(R.begin(), R.end());
  for (int x : R) {
    cout << x << ' ';
  }
  return 0;
}
