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 
#include <bits/stdc++.h>
using namespace std;
const int N = 200100;
struct node {
  double r, s;
  bool operator<(const node& a) const { return r == a.r ? s > a.s : r > a.r; }
} a[N], d[N];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    d[i] = a[i] = {(double)x, (double)y};
  }
  sort(a, a + n);
  vector<node> v, p;
  for (int i = 0, s = 0, r = 0; i < n; i++) {
    if (a[i].s > s) s = a[i].s, p.push_back(a[i]);
  }
  for (int i = p.size() - 1; i >= 0; i--) {
    function<bool(node, node, node)> cmp = [](node a, node b, node c) {
      function<double(node, node)> slope = [](node a, node b) {
        return (b.r - a.r) / (a.r * b.r) / ((a.s - b.s) / (a.s * b.s));
      };
      return slope(a, b) < slope(b, c);
    };
    while (v.size() > 1 && cmp(v.end()[-2], v.end()[-1], p[i])) v.pop_back();
    v.push_back(p[i]);
  }
  set<node> s;
  for (auto d : v) s.insert(d);
  for (int i = 0; i < n; i++)
    if (s.count(d[i])) printf("%d ", i + 1);
  puts("");
  return 0;
}
