Tavas is a cheerleader in the new sports competition named "Pashmaks".

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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;
int n, m;
int dcmp(double x) { return (x > 1e-10) - (x < -1e-10); }
struct Point {
  double x, y;
  Point() {}
  Point(double x, double y) : x(x), y(y) { id = 0; }
  int id;
  bool operator<(const Point &other) const {
    return dcmp(x - other.x) < 0 ||
           (dcmp(x - other.x) == 0 && dcmp(y - other.y) < 0);
  }
} p[200010], ch[200010];
int out[200010];
Point operator-(Point A, Point B) { return Point(A.x - B.x, A.y - B.y); }
double Cross(Point A, Point B) { return A.x * B.y - A.y * B.x; }
int ConvexHull(Point *p, int n, Point *ch) {
  sort(p, p + n);
  int m = 0;
  for (int i = 0; i < n; ++i) {
    while (m > 1 && Cross(ch[m - 1] - ch[m - 2], p[i] - ch[m - 2]) < 0) m--;
    ch[m++] = p[i];
  }
  return m;
}
int main() {
  int i, j;
  scanf("%d", &n);
  double y = 1000000000.0;
  for (i = 0; i <= n; ++i) {
    scanf("%lf%lf", &p[i].x, &p[i].y);
    p[i].x = 10000.0 / p[i].x;
    p[i].y = 10000.0 / p[i].y;
    y = min(p[i].y, y);
    p[i].id = i;
  }
  m = ConvexHull(p, n, ch);
  int cnt = 0;
  for (i = 0; i < m; ++i) {
    out[cnt++] = ch[i].id;
    if (fabs(y - ch[i].y) < 1e-10) break;
  }
  sort(out, out + cnt);
  for (i = 0; i < cnt; ++i) printf("%d ", out[i] + 1);
  printf("\n");
}
