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 int MAXN = 200005;
long long ccw(pair<long long, long long> a, pair<long long, long long> b,
              pair<long long, long long> c) {
  long long ret = 0;
  ret += b.first * c.first * a.second * (c.second - b.second);
  ret += c.first * a.first * b.second * (a.second - c.second);
  ret += a.first * b.first * c.second * (b.second - a.second);
  return ret;
}
map<pair<long long, long long>, int> M;
int r[MAXN];
int s[MAXN];
pair<long long, long long> p[MAXN];
pair<long long, long long> h[MAXN];
int main(void) {
  int n;
  scanf("%d", &n);
  set<pair<long long, long long> > S;
  for (int i = (0); i < (n); ++i) {
    scanf("%d%d", &r[i], &s[i]);
    S.insert({r[i], s[i]});
  }
  int m = 0;
  for (auto it : S) p[m++] = it;
  auto cmp = [&](pair<long long, long long> a, pair<long long, long long> b) {
    if (a.first != b.first) return a.first > b.first;
    return a.second < b.second;
  };
  sort(p, p + m, cmp);
  int top = 0;
  for (int i = (0); i < (m); ++i) {
    while (top >= 2 && ccw(h[top - 2], h[top - 1], p[i]) < 0) --top;
    h[top++] = p[i];
  }
  while (top >= 2 && h[top - 1].second <= h[top - 2].second) --top;
  int start = 0;
  while (start + 1 < top && h[start].first == h[start + 1].first) ++start;
  for (int i = (start); i < (top); ++i) M[h[i]] = 1;
  for (int i = (0); i < (n); ++i)
    if (M[{r[i], s[i]}]) printf("%d ", i + 1);
  puts("");
  return 0;
}
