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 Car {
  long double a, b;
  int num;
  bool operator<(const Car &p) const {
    if (a == p.a) return b < p.b;
    return a < p.a;
  }
} c[200005];
long double jd(int i, int j) { return (c[j].b - c[i].b) / (c[i].a - c[j].a); }
int dcmp(long double a, long double b) {
  if (fabs(a - b) < 1e11)
    return 0;
  else if (a < b)
    return -1;
  else
    return 1;
}
int s[200005], top = 0, ans[200005];
int main() {
  int n, i;
  double a, b;
  long double H = 1;
  cin >> n;
  for (i = 1; i <= n; i++) {
    scanf("%lf%lf", &a, &b);
    c[i].b = -(H / a);
    c[i].a = (long double)(b - a) / (long double)(a * b);
    c[i].num = i;
  }
  sort(c + 1, c + 1 + n);
  for (i = 1; i <= n; i++) {
    if (i != n && dcmp(c[i].a, c[i + 1].a) == 0 && c[i].b < c[i + 1].b)
      continue;
    while (top > 0 && jd(i, s[top]) < -(1e-11)) top--;
    while (top > 1 && jd(s[top], s[top - 1]) - jd(i, s[top]) > 1e-11) top--;
    s[++top] = i;
  }
  while (top > 1 && dcmp(jd(s[top - 1], s[top]), -1) == -1) top--;
  for (i = 1; i <= top; i++) ans[i] = c[s[i]].num;
  sort(ans + 1, ans + 1 + top);
  for (i = 1; i < top; i++) printf("%d ", ans[i]);
  printf("%d", ans[i]);
  return 0;
}
