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 point {
  double x, y, z;
  inline bool operator<(const point t) const { return y * t.x - x * t.y > 0; }
  inline bool operator>(const point t) const { return y * t.x - x * t.y < 0; }
  inline bool operator==(const point t) const { return y * t.x - x * t.y == 0; }
  inline point operator-(const point t) const {
    return (point){x - t.x, y - t.y, 0};
  }
};
point p[300000];
vector<int> V[300000];
int q[300000];
int i, j, n, o, t;
double mx, my;
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++)
    scanf("%lf%lf", &p[i].x, &p[i].y), p[i].x = 1 / p[i].x, p[i].y = 1 / p[i].y,
                                       p[i].z = i;
  sort(p + 1, p + n + 1);
  p[0] = (point){0, 100000, 0}, p[n + 1] = (point){100000, 0, 0};
  for (i = 0; i <= n + 1; i++) {
    if ((i) && (p[i].x == p[q[t]].x) && (p[i].y == p[q[t]].y)) {
      V[t].push_back(p[i].z);
      continue;
    }
    for (; (t > 1) && (p[i] - p[q[t]] > p[q[t]] - p[q[t - 1]]); t--)
      ;
    t++, V[t].clear(), q[t] = i, V[t].push_back(p[i].z);
  }
  for (i = 1; i <= t; i++)
    for (j = 0; j < V[i].size(); j++)
      if (V[i][j]) o++, q[o] = V[i][j];
  sort(q + 1, q + o + 1);
  for (i = 1; i <= o; i++) printf("%d ", q[i]);
  return 0;
}
