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;
template <class T>
T abs(T x) {
  return x > 0 ? x : -x;
}
const int N = 200000;
int n;
int m;
pair<pair<int, int>, int> v[N];
int x[N];
int y[N];
vector<pair<int, int> > w[N];
int bad[N];
int q[N + 10];
bool l1(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  if (a.first.first != b.first.first) return a.first.first < b.first.first;
  if (a.first.second != b.first.second) return a.first.second > b.first.second;
  return a.second < b.second;
}
long long vect(int a, int b, int c) {
  long long xa = v[a].first.first;
  long long ya = v[a].first.second;
  long long xb = v[b].first.first;
  long long yb = v[b].first.second;
  long long xc = v[c].first.first;
  long long yc = v[c].first.second;
  return xc * yb * (xa - xb) * (ya - yc) - xb * yc * (xa - xc) * (ya - yb);
}
long long vect(long long xa, long long ya, long long xb, long long yb,
               long long xc, long long yc) {
  return xc * yb * (xa - xb) * (ya - yc) - xb * yc * (xa - xc) * (ya - yb);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &v[i].first.first, &v[i].first.second);
    x[i] = v[i].first.first;
    y[i] = v[i].first.second;
    v[i].second = i;
  }
  sort(v, v + n);
  int top = -1;
  for (int i = n - 1; i >= 0; i--) {
    if (top >= v[i].first.second) bad[v[i].second] = 1;
    if (i == 0 || v[i - 1].first != v[i].first)
      top = max(top, v[i].first.second);
  }
  {
    sort(v, v + n, l1);
    int o = 0;
    for (int i = 0; i < n; i++) {
      while (o > 1 && vect(i, q[o - 2], q[o - 1]) >= 0) o--;
      if (o > 0) {
        w[v[i].second].push_back(v[q[o - 1]].first);
      }
      q[o++] = i;
    }
  }
  for (int i = 0; i < n; i++) swap(v[i].first.first, v[i].first.second);
  {
    sort(v, v + n, l1);
    int o = 0;
    for (int i = 0; i < n; i++) {
      while (o > 1 && vect(i, q[o - 2], q[o - 1]) >= 0) o--;
      if (o > 0) {
        w[v[i].second].push_back(
            make_pair(v[q[o - 1]].first.second, v[q[o - 1]].first.first));
      }
      q[o++] = i;
    }
  }
  for (int i = 0; i < n; i++)
    if (((int)(w[i]).size()) == 2 &&
        vect(x[i], y[i], w[i][0].first, w[i][0].second, w[i][1].first,
             w[i][1].second) < 0)
      bad[i] = 1;
  for (int i = 0; i < n; i++)
    if (!bad[i]) printf("%d ", i + 1);
  printf("\n");
  return 0;
}
