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;
int n;
map<pair<int, int>, vector<int> > M;
pair<int, int> a[222222];
vector<int> ans;
bool bad(int a, int b, int c, int d, int e, int first) {
  long long xx = 1LL * (b - d) * (e - c) * a * first;
  long long yy = 1LL * (d - first) * (c - a) * b * e;
  return xx > yy;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    M[make_pair(x, y)].push_back(i);
    a[i] = make_pair(x, y);
  }
  sort(a, a + n);
  n = unique(a, a + n) - a;
  int an = 0;
  for (int i = 0; i < n; i++) {
    while (an > 0 && a[an - 1].second <= a[i].second) {
      --an;
    }
    a[an++] = a[i];
  }
  n = an;
  reverse(a, a + n);
  an = 0;
  for (int i = 0; i < n; i++) {
    while (an > 1 && bad(a[i].first, a[i].second, a[an - 1].first,
                         a[an - 1].second, a[an - 2].first, a[an - 2].second)) {
      --an;
    }
    a[an++] = a[i];
  }
  n = an;
  for (int i = 0; i < n; i++) {
    for (__typeof(M[a[i]].begin()) it = M[a[i]].begin(); it != M[a[i]].end();
         ++it) {
      ans.push_back(*it);
    }
  }
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); i++) {
    if (i > 0) {
      putchar(' ');
    }
    printf("%d", ans[i] + 1);
  }
  puts("");
  return 0;
}
