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;
  int id;
  bool operator<(point a) const { return x != a.x ? x < a.x : y < a.y; }
} p[200010];
int n, tot;
int s[200010], top;
std::vector<int> ans, vec[200010];
inline double calc(int a, int b) {
  return (p[a].y - p[b].y) / (p[a].x - p[b].x);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int r, s;
    scanf("%d%d", &r, &s);
    p[i] = (point){(double)1.0 / r, (double)1.0 / s, i};
  }
  sort(p + 1, p + 1 + n);
  tot = 1;
  for (int i = 2; i <= n; i++)
    if (p[i].x != p[tot].x)
      p[++tot] = p[i];
    else if (p[i].y == p[tot].y)
      vec[p[tot].id].push_back(p[i].id);
  s[1] = 1;
  top = 1;
  for (int i = 2; i <= tot; i++) {
    while (top > 1 && calc(s[top - 1], s[top]) > calc(s[top - 1], i)) top--;
    s[++top] = i;
  }
  ans.push_back(p[s[1]].id);
  for (auto i : vec[p[s[1]].id]) ans.push_back(i);
  for (int i = 2; i <= top; i++) {
    if (calc(s[i - 1], s[i]) < 0) {
      ans.push_back(p[s[i]].id);
      for (int j : vec[p[s[i]].id]) ans.push_back(j);
    }
  }
  sort(ans.begin(), ans.end());
  for (auto i : ans) printf("%d ", i);
  return 0;
}
