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;
const int N = 2e5 + 5;
const double eps = 1e-9;
int n;
struct node {
  double x, y;
  int id;
  node(double rx = 0, double ry = 0, int ri = 0) {
    x = rx;
    y = ry;
    id = ri;
  }
  bool operator<(const node& rhs) const {
    return x != rhs.x ? x < rhs.x : y < rhs.y;
  }
  node operator-(const node& rhs) const {
    return node(x - rhs.x, y - rhs.y, id);
  }
  double operator^(const node& rhs) const { return x * rhs.y - y * rhs.x; }
} nd[N];
node stdw[N];
double cross(node u, node v, node w) { return (v - u) ^ (w - v); }
bool idcmp(node u, node v) { return u.id < v.id; }
int tpdw;
bool dwempty() { return tpdw < 2; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    static double u, v;
    scanf("%lf%lf", &u, &v);
    nd[i] = node(1.0 / u, 1.0 / v, i);
  }
  sort(nd + 1, nd + n + 1);
  stdw[++tpdw] = nd[1];
  for (int i = 2; i <= n; i++) {
    if (!dwempty() && cross(stdw[tpdw - 1], stdw[tpdw], nd[i]) < eps) {
      --tpdw;
    }
    stdw[++tpdw] = nd[i];
  }
  while (!dwempty() && cross(node(-1e5, stdw[tpdw - 1].y, 0), stdw[tpdw - 1],
                             stdw[tpdw]) > -eps) {
    --tpdw;
  }
  sort(stdw + 1, stdw + tpdw + 1, idcmp);
  for (int i = 1; i <= tpdw; i++) {
    printf("%d%c", stdw[i].id, " \n"[i == tpdw]);
  }
  return 0;
}
