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;
struct pnt {
  int x, y, id;
  void in(int i) {
    scanf("%d %d", &x, &y);
    id = i;
  }
  bool operator<(const pnt &p) const { return x != p.x ? x > p.x : y > p.y; }
  bool operator==(const pnt &p) const { return x == p.x && y == p.y; }
} p[200100], q[200100];
long long cross(pnt a, pnt b, pnt c) {
  return 1ll * b.y * c.x * (a.x - b.x) * (a.y - c.y) -
         1ll * b.x * c.y * (a.y - b.y) * (a.x - c.x);
}
int n;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) p[i].in(i), q[i] = p[i];
  sort(p, p + n);
  int m = unique(p, p + n) - p;
  vector<pnt> rlt;
  vector<int> ans;
  for (int i = 0, j = 0; i < m; i++) {
    while (j > 1 && cross(rlt[j - 2], rlt[j - 1], p[i]) < 0)
      rlt.pop_back(), j--;
    rlt.push_back(p[i]), j++;
  }
  int k = adjacent_find(rlt.begin(), rlt.end(),
                        [&](pnt a, pnt b) { return a.y >= b.y; }) -
          rlt.begin();
  if (k != rlt.size()) k++;
  sort(q, q + n);
  for (int i = 0, j = 0; i < k; i++) {
    while (!(q[j] == rlt[i])) j++;
    while (j < n && q[j] == rlt[i]) ans.push_back(q[j].id), j++;
  }
  sort(ans.begin(), ans.end());
  for (auto x : ans) printf("%d ", x + 1);
  return 0;
}
