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;
void read(int &x) {
  x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9')
    x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
}
const long double eps = 1e-8;
int n, s[300010], tot;
struct Node {
  int a, b, id;
  bool operator<(const Node &x) const {
    return (a < x.a && b <= x.b) || (a <= x.a && b < x.b);
  }
  bool operator==(const Node &x) const { return a == x.a && b == x.b; }
} d[300010];
Node p[300010], q[300010];
long double slope(Node i, Node j) {
  return (i.b - j.b) * 1.0 / (i.a - j.a) / i.b / j.b * i.a * j.a;
}
inline int cmp(Node x, Node y) {
  return x.a ^ y.a ? x.a > y.a : x.b ^ y.b ? x.b > y.b : x.id < y.id;
}
vector<int> v[300010];
bool acc[300010];
int main() {
  read(n);
  for (register int i = 1; i <= n; ++i)
    read(d[i].a), read(d[i].b), d[i].id = i, acc[i] = true;
  sort(d + 1, d + n + 1, cmp);
  for (register int i = 1; i <= n; ++i) {
    if (!tot)
      s[++tot] = i, p[tot] = (Node){d[i].a, d[i].b, d[i].id};
    else {
      int u = s[tot];
      if (d[i] == d[u])
        v[d[u].id].push_back(i), acc[d[i].id] = false;
      else if (d[i] < d[u])
        acc[d[i].id] = false;
      else
        s[++tot] = i, p[tot] = (Node){d[i].a, d[i].b, d[i].id};
    }
  }
  int head = 0, tail = 1;
  q[1] = p[1];
  for (register int i = 2; i <= tot; ++i) {
    while (head + 1 < tail &&
           slope(q[tail - 1], q[tail]) > slope(q[tail - 1], p[i]))
      acc[q[tail].id] = false, --tail;
    q[++tail] = p[i];
  }
  for (register int i = 1; i <= n; ++i)
    if (acc[i]) {
      printf("%d ", i);
      int tot = v[i].size();
      for (int j = 0; j ^ tot; ++j) acc[d[v[i][j]].id] = true;
    }
}
