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 Girl {
  double s, r;
};
Girl girl[200000];
int p[200000];
int shell[200000];
int all[200000];
int num = 0;
int total = 0;
bool cmp(const int& a, const int& b) {
  return girl[a].s < girl[b].s ||
         (girl[a].s == girl[b].s && girl[a].r > girl[b].r);
}
Girl operator-(const Girl& a, const Girl& b) {
  return (Girl){a.s - b.s, a.r - b.r};
}
double cross(const Girl& a, const Girl& b) { return a.s * b.r - a.r * b.s; }
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%lf %lf", &girl[i].s, &girl[i].r);
    girl[i].s = 1 / girl[i].s;
    girl[i].r = 1 / girl[i].r;
    p[i] = i;
  }
  sort(p, p + n, cmp);
  for (int i = 0; i < n; ++i) {
    if (!(num > 0 && ((girl[p[i]].s > girl[shell[num - 1]].s &&
                       girl[p[i]].r >= girl[shell[num - 1]].r) ||
                      (girl[p[i]].s >= girl[shell[num - 1]].s &&
                       girl[p[i]].r > girl[shell[num - 1]].r)))) {
      if (!(num > 0 && girl[p[i]].s == girl[shell[num - 1]].s &&
            girl[p[i]].r == girl[shell[num - 1]].r)) {
        while ((num > 0 && girl[p[i]].s <= girl[shell[num - 1]].s &&
                girl[p[i]].r <= girl[shell[num - 1]].r) ||
               (num > 1 && cross(girl[shell[num - 1]] - girl[shell[num - 2]],
                                 girl[p[i]] - girl[shell[num - 2]]) <= 0)) {
          while (total > 0 &&
                 girl[all[total - 1]].s == girl[shell[num - 1]].s &&
                 girl[all[total - 1]].r == girl[shell[num - 1]].r) {
            total--;
          }
          num--;
        }
        shell[num++] = p[i];
      }
      all[total++] = p[i];
    }
  }
  sort(all, all + total);
  for (int i = 0; i < total; ++i) {
    if (i) {
      printf(" ");
    }
    printf("%d", all[i] + 1);
  }
  printf("\n");
  return 0;
}
