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 {
  long long x, y;
};
long long cross(const Point &a, const Point &b, const Point &c) {
  return a.y * c.x * (a.x - b.x) * (b.y - c.y) -
         a.x * c.y * (a.y - b.y) * (b.x - c.x);
}
Point p[10004];
Point cv[10004];
vector<int> id[10004];
int ans[200004], ansTop(-1);
void add(vector<int> &all) {
  for (int i = 0; i < all.size(); ++i) ans[++ansTop] = all[i];
}
int main() {
  for (int i = 1; i <= 10000; ++i) p[i].x = i;
  int n, maxY;
  scanf(" %d", &n);
  for (int i = 1, x, y; i <= n; ++i) {
    scanf(" %d%d", &x, &y);
    if (p[x].y == y)
      id[x].push_back(i);
    else if (p[x].y < y) {
      p[x].y = y;
      id[x].clear();
      id[x].push_back(i);
    }
    if (y > maxY) maxY = y;
  }
  int top = -1;
  for (int i = 10000; i > 0; --i) {
    if (p[i].y == 0LL) continue;
    while (top > 0 && cross(cv[top - 1], cv[top], p[i]) < 0) --top;
    cv[++top] = p[i];
    if (p[i].y == maxY) break;
  }
  add(id[cv[0].x]);
  for (int i = 1; i <= top; ++i) {
    add(id[cv[i].x]);
    if (cv[i].y == maxY) break;
  }
  sort(&ans[0], &ans[ansTop + 1]);
  for (int i = 0; i <= ansTop; ++i) printf("%d ", ans[i]);
  printf("\n");
}
