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 long long int INF = (long long int)1e9 + 10;
const long long int INFLL = (long long int)1e18 + 10;
const long double EPS = 1e-8;
const long double EPSLD = 1e-18;
const long long int MOD = 1e9 + 7;
template <class T>
T &chmin(T &a, const T &b) {
  return a = min(a, b);
}
template <class T>
T &chmax(T &a, const T &b) {
  return a = max(a, b);
}
long long int cross(pair<long long int, long long int> a,
                    pair<long long int, long long int> b) {
  return a.first * b.second - a.second * b.first;
}
long long int ccw(pair<long long int, long long int> a,
                  pair<long long int, long long int> b,
                  pair<long long int, long long int> c) {
  pair<long long int, long long int> nb = pair<long long int, long long int>(
      (a.first - b.first) * c.first * b.second, a.second - b.second);
  pair<long long int, long long int> nc = pair<long long int, long long int>(
      (a.first - c.first) * b.first * c.second, a.second - c.second);
  if (cross(nb, nc) > 0) return +1;
  if (cross(nb, nc) < 0) return -1;
  return 0;
}
const long long int MAX_N = 200010;
long long int n;
pair<pair<long long int, long long int>, long long int> v[MAX_N];
pair<pair<long long int, long long int>, long long int> ch[MAX_N];
vector<long long int> ans;
int main() {
  scanf("%lld", &n);
  for (long long int i = (0); i < (long long int)(n); i++)
    scanf("%lld %lld", &v[i].first.first, &v[i].first.second);
  for (long long int i = (0); i < (long long int)(n); i++) v[i].second = i;
  sort(v, v + n);
  reverse(v, v + n);
  long long int k = 0;
  for (long long int i = 0; i < n; ch[k++] = v[i++])
    while (k >= 2 && ccw(ch[k - 2].first, ch[k - 1].first, v[i].first) == -1)
      k--;
  for (long long int i = (0); i < (long long int)(k); i++) {
    if (i > 0 && ch[i].first.second <= ch[i - 1].first.second) continue;
    ans.emplace_back(ch[i].second);
  }
  sort((ans).begin(), (ans).end());
  for (long long int i = (0);
       i < (long long int)(((long long int)(ans).size())); i++)
    printf("%lld%c", ans[i] + 1,
           i == ((long long int)(ans).size()) - 1 ? '\n' : ' ');
  return 0;
}
