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;
const int _ = 1e6 + 20;
double X, Y;
struct point {
  double x, y;
  int id;
  double slope() { return y / x; }
  point friend operator-(point x, point y) {
    return (point){x.x - y.x, x.y - y.y};
  }
  int friend operator==(point x, point y) { return abs(x.y - y.y) < 1e-9; }
} a[_], q[_];
int cmp(point x, point y) {
  if (x.x == y.x) return x.y > y.y;
  return x.x < y.x;
}
int n, p[_], fa[_], k = 1, top = 1;
int pd(point x, point y, point z) { return (y - x).slope() > (z - y).slope(); }
int fi(int x) { return fa[x] ? fa[x] = fi(fa[x]) : x; }
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int x, y;
    cin >> x >> y;
    a[i].x = (double)1 / x;
    a[i].y = (double)1 / y;
    a[i].id = i;
  }
  sort(a + 1, a + n + 1, cmp);
  while (a[k].x == a[k + 1].x && k < n) {
    if (a[k].y == a[k + 1].y) fa[a[k].id] = a[k + 1].id;
    ++k;
  }
  X = a[k].x;
  Y = a[k].y;
  q[1] = a[k];
  for (int i = k + 1; i <= n; ++i) {
    while (a[i].x == a[i + 1].x) {
      if (a[i].y == a[i + 1].y) fa[a[i].id] = a[i + 1].id;
      ++i;
    }
    while (top > 1 && pd(q[top - 1], q[top], a[i])) --top;
    q[++top] = a[i];
  }
  for (int i = 1; i <= top; ++i) {
    if (i != 1 && (q[i].y >= q[i - 1].y)) break;
    p[q[i].id] = 1;
  }
  for (int i = 1; i <= n; ++i)
    if (p[fi(i)]) cout << i << " ";
  cout << endl;
  return 0;
}
