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 N = 101010;
int n, m = 1e4;
bool ans[N];
pair<pair<int, int>, int> a[N];
struct Seg {
  static const int N = ::N;
  int cnt[N];
  void upd(int p, int l, int r, int rt) {
    ++cnt[rt];
    if (l == r) return;
    int mid = l + r >> 1;
    if (p <= mid)
      upd(p, l, mid, rt << 1);
    else
      upd(p, mid + 1, r, rt << 1 | 1);
  }
  int qry(int L, int R, int l, int r, int rt) {
    if (L > R) return 0;
    if (L <= l && r <= R) return cnt[rt];
    int mid = l + r >> 1, ans = 0;
    if (L <= mid) ans += qry(L, R, l, mid, rt << 1);
    if (R >= mid + 1) ans += qry(L, R, mid + 1, r, rt << 1 | 1);
    return ans;
  }
} seg;
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  cin >> n;
  for (int i = (1); i < (n + 1); i++)
    cin >> a[i].first.first >> a[i].first.second, a[i].second = i;
  sort(a + 1, a + 1 + n);
  reverse(a + 1, a + 1 + n);
  for (int i = (1); i < (n + 1); i++) {
    int x = a[i].first.first, y = a[i].first.second;
    if (seg.qry(a[i].first.second + 1, m, 1, m, 1) == 0) ans[a[i].second] = 1;
    seg.upd(a[i].first.second, 1, m, 1);
  }
  for (int i = (1); i < (n + 1); i++)
    if (ans[i]) cout << i << " ";
  cout << "\n";
  return 0;
}
