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;
int a[200000], b[200000];
double fa[200000], fb[200000];
int pos[11000];
bool bad[11000];
int main() {
  scanf("%d", &N);
  memset((pos), (-1), sizeof(pos));
  for (int(i) = 0; (i) < (N); (i)++) {
    scanf("%d%d", &a[i], &b[i]);
    if (pos[a[i]] == -1 || b[i] > b[pos[a[i]]]) pos[a[i]] = i;
    fa[i] = 1.0 / a[i];
    fb[i] = 1.0 / b[i];
  }
  vector<int> lis;
  for (int(i) = 0; (i) < (11000); (i)++)
    if (pos[i] != -1) lis.push_back(pos[i]);
  vector<int> ans;
  for (int(i) = 0; (i) < (lis.size()); (i)++) {
    double lb = 1e-7, ub = 1e8;
    for (int(j) = 0; (j) < (lis.size()); (j)++)
      if (i != j) {
        int p1 = lis[i], p2 = lis[j];
        double k1 = fa[p1] - fa[p2];
        double k2 = fb[p2] - fb[p1];
        if (k1 < 0)
          lb = max(lb, k2 / k1);
        else
          ub = min(ub, k2 / k1);
      }
    if (ub - lb > -1e-7) ans.push_back(lis[i] + 1);
  }
  sort((ans).begin(), (ans).end());
  for (int(i) = 0; (i) < (ans.size()); (i)++)
    printf("%d%c", ans[i], i + 1 == ans.size() ? '\n' : ' ');
  ;
}
