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 N = 200010;
struct pr {
  int x, y, ind;
};
pr t, v[N], s[N];
int n, top, maxx = 1, maxy = 1, minx = 1;
bool ans[N];
inline bool xi(pr x, pr y) { return x.x < y.x || x.x == y.x && x.y < y.y; }
inline int di(pr x, pr y, pr z) {
  return (y.x - x.x) * (z.y - x.y) - (z.x - x.x) * (y.y - x.y);
}
bool cmp(pr x, pr y) {
  int temp = di(v[1], x, y);
  if (temp > 0 || temp == 0 && xi(x, y)) return 1;
  return 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &v[i].x, &v[i].y);
    v[i].ind = i;
  }
  sort(v + 2, v + n + 1, cmp);
  s[++top] = v[1];
  s[++top] = v[2];
  for (int i = 3; i <= n; i++) {
    while (di(s[top - 1], s[top], v[i]) < 0) top--;
    s[++top] = v[i];
  }
  for (int i = 1; i <= top; i++) {
    if (s[i].x > s[maxx].x || s[i].x == s[maxx].x && s[i].y > s[maxx].y)
      maxx = i;
    if (s[i].y > s[maxy].y || s[i].y == s[maxy].y && s[i].x > s[maxx].x)
      maxy = i;
  }
  for (int i = 1; i <= top; i++)
    if (s[i].y > s[maxx].y && s[i].x > s[maxy].x ||
        s[i].x == s[maxx].x && s[i].y == s[maxx].y ||
        s[i].x == s[maxy].x && s[i].y == s[maxy].y)
      ans[s[i].ind] = 1;
  for (int i = 1; i <= n; i++)
    if (ans[i]) printf("%d ", i);
  return 0;
}
