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 double eps = 1e-8;
int n;
bool vis[220000];
int dcmp(double x) {
  if (fabs(x) < eps) return 0;
  return x < 0 ? -1 : 1;
}
map<pair<double, double>, int> M;
struct Point {
  double x, y;
  int id;
  Point() {}
  Point(double _x, double _y, int _z) : x(_x), y(_y), id(_z) {}
  Point operator-(const Point& b) const { return Point(x - b.x, y - b.y, 0); }
  double operator*(const Point& b) const { return (x * b.y - y * b.x); }
  double operator^(const Point& b) const { return (x * b.x + y * b.y); }
  bool operator<(const Point& b) const {
    if (dcmp(x - b.x) == 0) return y < b.y;
    return x < b.x;
  }
} p[220000], stk[220000];
int Graham(int n) {
  sort(p, p + n);
  int top = 0;
  for (int i = 0; i < n; i++) {
    if (i && !dcmp(p[i].x - p[i - 1].x)) continue;
    while (top > 1 && (stk[top - 1] - stk[top - 2]) * (p[i] - stk[top - 1]) < 0)
      top--;
    stk[top++] = p[i];
  }
  return top;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    double x, y;
    scanf("%lf%lf", &x, &y);
    x = 10000 / x;
    y = 10000 / y;
    p[i] = {x, y, i};
  }
  int top = Graham(n);
  for (int i = 0; i < top; i++) {
    if (!i || dcmp(stk[i].y - stk[i - 1].y) < 0) vis[stk[i].id] = 1;
  }
  for (int i = 0; i < n; i++) {
    if (vis[i]) printf("%d ", i + 1);
  }
  return 0;
}
