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>
const double eps = 1e-5;
using namespace std;
int n, tail;
bool vis[200010];
int f[200010];
struct point {
  int num;
  int x, y;
  void init(int num) {
    this->num = num;
    scanf("%d %d", &x, &y);
  }
} p[200010], *q[200010];
bool cmp_point(const point &a, const point &b) {
  if (a.x != b.x) return a.x > b.x;
  return a.y > b.y;
}
bool cmp(point *a, point *b, point *c) {
  long long A = (long long)b->x * c->y;
  long long B = (long long)c->x * b->y;
  return (long long)(a->x - b->x) * (a->y - c->y) * B <
         (long long)(a->x - c->x) * (a->y - b->y) * A;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    point *nt = p + i;
    nt->init(i);
  }
  sort(p + 1, p + n + 1, cmp_point);
  for (int i = 1; i <= n; ++i) {
    point *nt = p + i;
    if (i > 1 && p[i].x == p[i - 1].x && p[i].y == p[i - 1].y) {
      f[i] = f[i - 1];
      continue;
    }
    f[i] = i;
    while (tail >= 2 && cmp(q[tail - 1], q[tail], nt)) --tail;
    q[++tail] = nt;
  }
  for (int i = 1; i <= tail; ++i) {
    if (i > 1 && q[i]->y <= q[i - 1]->y) break;
    vis[q[i]->num] = true;
  }
  for (int i = 1; i <= n; ++i) {
    if (vis[p[f[i]].num] == true) vis[p[i].num] = true;
  }
  for (int i = 1; i <= n; ++i)
    if (vis[i]) printf("%d ", i);
}
