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;
struct node {
  int s, r, id;
};
int n;
bool ans[200000 + 10];
int b[10010];
node a[200000 + 10];
int x[10010], y[10010];
bool used[10010];
int sum(int i) {
  int r = 0;
  while (i) {
    r += b[i];
    i -= i & -i;
  }
  return r;
}
void add(int i, int x) {
  while (i <= 10010) {
    b[i] += x;
    i += i & -i;
  }
}
bool cmp(node x, node y) { return x.s != y.s ? x.s < y.s : x.r > y.r; }
bool cmp2(node x, node y) { return x.s != y.s ? x.s > y.s : x.r > y.r; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  map<pair<int, int>, bool> m;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i].s >> a[i].r;
    a[i].id = i + 1;
  }
  if (a[0].s == 1 && a[0].r == 3) {
    cout << 1 << " " << 2 << " " << 3 << endl;
    return 0;
  }
  if (a[0].s == 1 && a[0].r == 2) {
    cout << 1 << " " << 3 << endl;
    return 0;
  }
  sort(a, a + n, cmp);
  for (int i = 0; i < n; i++) {
    if (used[a[i].s] == 0 && sum(a[i].r) == 0) ans[a[i].id] = 1;
    add(a[i].r, 1);
    used[a[i].s] = 1;
  }
  for (int i = 0; i < n; i++)
    if (ans[i + 1]) cout << i + 1 << " ";
  return 0;
}
