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;
double intersect(pair<double, double> a, pair<double, double> b, double pv) {
  if (a.first == b.first) {
    if (a.second < b.second)
      return -1;
    else
      return pv;
  }
  return (b.second - a.second) / (a.first - b.first);
}
int main() {
  int n;
  scanf("%d", &n);
  vector<pair<pair<double, double>, int> > points(n);
  for (int i = 0; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    points[i].first.first = 1.0 / x - 1.0 / y;
    points[i].first.second = 1.0 / y;
    points[i].second = i + 1;
  }
  sort(points.begin(), points.end());
  stack<pair<double, int> > st;
  for (int i = n - 1; i >= 0; i--) {
    double inter = 0.0;
    while (!st.empty()) {
      pair<double, int> P = st.top();
      inter = intersect(points[i].first, points[P.second].first, P.first);
      if (inter < P.first ||
          (inter <= 0 && points[i].first != points[P.second].first)) {
        st.pop();
      } else {
        break;
      }
    }
    if (st.empty()) {
      st.push(make_pair(0, i));
    } else if (1 - inter >= 0.0000000000001) {
      st.push(make_pair(inter, i));
    }
  }
  vector<int> r;
  while (!st.empty()) {
    if (st.top().first >= 1) {
      continue;
    } else {
      r.push_back(points[st.top().second].second);
    }
    st.pop();
  }
  sort(r.begin(), r.end());
  for (int i = 0; i < r.size(); i++) {
    printf("%d ", r[i]);
  }
  printf("\n");
}
