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 
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 100;
map<pair<int, int>, vector<int> > mp;
const long double eps = 1e-22;
struct point {
  int X, Y;
  long double first, second;
  point() { X = Y = first = second = .0; }
  point(long double a, long double b) { first = a, second = b; }
  inline void init(int a, int b) {
    X = a, Y = b;
    first = 1.0 / X;
    second = 1.0 / Y;
  }
  inline long double operator*(point p) {
    return first * p.second - second * p.first;
  }
  inline point operator-(point p) {
    return point(first - p.first, second - p.second);
  }
  inline long double dis(point p) {
    return (((first - p.first) * (first - p.first))) +
           (((second - p.second) * (second - p.second)));
  }
  inline pair<int, int> PII() { return pair<int, int>(X, Y); }
};
point o;
point l;
inline bool ocmp(const point &a, const point &b) {
  return make_pair(a.second, a.first) < make_pair(b.second, b.first);
}
inline bool lcmp(const point &a, const point &b) {
  return make_pair(a.first, a.second) < make_pair(b.first, b.second);
}
inline bool scmp(const point &a, const point &b) {
  long double cross = ((point)a - o) * ((point)b - o);
  if (fabs(cross) <= eps) return o.dis(a) < o.dis(b);
  return cross < eps;
}
vector<point> v;
vector<point> hull;
int n;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int(i) = (0); (i) < (n); ++(i)) {
    int a, b;
    cin >> a >> b;
    mp[{a, b}].push_back(i + 1);
  }
  if (mp.size() == 1) {
    for (int(i) = (1); (i) < (n + 1); ++(i)) cout << i << ' ';
    cout << endl;
    return 0;
  }
  for (__typeof((mp).begin()) q = (mp).begin(); q != (mp).end(); ++q) {
    pair<int, int> p = q->first;
    int first = p.first, second = p.second;
    point P;
    P.init(first, second);
    v.push_back(P);
  }
  o = *min_element(v.begin(), v.end(), ocmp);
  l = *min_element(v.begin(), v.end(), lcmp);
  if (o.PII() == l.PII()) {
    for (auto &(a) : (mp[o.PII()])) cout << a << ' ';
    cout << endl;
    return 0;
  }
  sort(v.begin(), v.end(), scmp);
  int po = 2;
  for (int(i) = (0); (i) < (2); ++(i)) hull.push_back(v[i]);
  vector<int> ans;
  while (po < v.size() && hull.back().PII() != l.PII()) {
    while (hull.size() > 1 && (hull[hull.size() - 1] - hull[hull.size() - 2]) *
                                      (v[po] - hull[hull.size() - 2]) >
                                  eps)
      hull.pop_back();
    hull.push_back(v[po++]);
  }
  for (auto &(p) : (hull))
    for (auto &(a) : (mp[p.PII()])) ans.push_back(a);
  sort(ans.begin(), ans.end());
  for (auto &(a) : (ans)) cout << a << ' ';
  cout << endl;
}
