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>
#pragma comment(linker, "/STACK:67108864")
using namespace std;
const int maxn = 2e5 + 100;
int n;
pair<int, int> p[maxn];
int r[maxn];
bool win[maxn];
pair<long long, long long> intersect(int a, int b) {
  pair<long long, long long> ans;
  ans.first = r[a] - r[b];
  ans.first *= b;
  ans.second = b - a;
  ans.second *= r[b];
  return ans;
}
bool comp(pair<long long, long long> a, pair<long long, long long> b) {
  if ((a.second < 0) ^ (b.second < 0))
    return a.first * b.second >= b.first * a.second;
  return a.first * b.second <= b.first * a.second;
}
bool is_bad(pair<long long, long long> p) { return p.first * p.second <= 0; }
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &p[i].first, &p[i].second);
    r[p[i].first] = max(r[p[i].first], p[i].second);
  }
  vector<int> lines;
  for (int i = 1; i <= 10000; i++) {
    if (r[i] == 0) continue;
    while (lines.size() > 1) {
      pair<long long, long long> p1 =
          intersect(lines[lines.size() - 1], lines[lines.size() - 2]);
      pair<long long, long long> p2 = intersect(lines[lines.size() - 1], i);
      if (comp(p1, p2)) break;
      lines.pop_back();
    }
    lines.push_back(i);
  }
  win[lines.back()] = true;
  for (int i = 0; i < lines.size() - 1; i++)
    if (!is_bad(intersect(lines[i], lines[i + 1]))) win[lines[i]] = true;
  for (int i = 0; i < n; i++) {
    if (win[p[i].first] && r[p[i].first] == p[i].second) printf("%d ", i + 1);
  }
  return 0;
}
