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;
vector<pair<pair<long long, long long>, int> > v, v1;
vector<pair<pair<long long, long long>, int> > cht;
map<long long, long long> my;
int c[200011] = {0};
bool intersection(int l1, int l2, int l3) {
  long long p1 = (cht[l2].first.second * cht[l3].first.first *
                  (cht[l3].first.second - cht[l1].first.second) *
                  (cht[l1].first.first - cht[l2].first.first));
  long long p2 = (cht[l3].first.second * cht[l2].first.first *
                  (cht[l2].first.second - cht[l1].first.second) *
                  (cht[l1].first.first - cht[l3].first.first));
  int p;
  if (p1 < p2)
    p = 1;
  else
    p = 0;
  return p;
}
void add(long long m, long long b, int i) {
  cht.push_back(make_pair(make_pair(m, b), i));
  while (cht.size() > 2 &&
         intersection(cht.size() - 3, cht.size() - 2, cht.size() - 1))
    cht.erase(cht.end() - 2);
}
int main() {
  int i, j, n, n1;
  cin >> n;
  n1 = n;
  for (i = 0; i < n; i++) {
    long long r, second;
    cin >> second >> r;
    v1.push_back(make_pair(make_pair(r, second), i + 1));
  }
  sort(v1.begin(), v1.end());
  for (i = 0; i < n; i++) {
    for (j = i; j < n; j++)
      if (v1[j].first.first != v1[i].first.first) break;
    v.push_back(
        make_pair(make_pair(v1[j - 1].first.first, v1[j - 1].first.second),
                  v1[j - 1].second));
    i = j - 1;
  }
  n = (int)v.size();
  add(v[0].first.first, v[0].first.second, v[0].second);
  for (i = 1; i < n; i++) {
    add(v[i].first.first, v[i].first.second, v[i].second);
  }
  for (i = 0; i < cht.size(); i++) {
    c[cht[i].second] = 1;
    my.insert(make_pair(cht[i].first.first, cht[i].first.second));
  }
  for (i = 0; i < n1; i++)
    if (c[i + 1] == 0 && my.find(v1[i].first.first) != my.end() &&
        my[v1[i].first.first] != v1[i].first.second)
      c[i + 1] = 1;
  for (i = 1; i <= n1; i++)
    if (c[i] == 1) printf("%d ", i);
  return 0;
}
