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;
using ll = long long;
struct point {
  ll x, y;
  bool operator<(point b) const {
    return make_pair(x, y) > make_pair(b.x, b.y);
  }
};
int n;
vector<point> a, p;
ll det(point a, point b, point c) {
  return (a.x - b.x) * (b.y - c.y) * c.x * a.y -
         (b.x - c.x) * (a.y - b.y) * a.x * c.y;
}
vector<point> CH;
void Push(point p) {
  while (CH.size() >= 2 && det(*++CH.rbegin(), *CH.rbegin(), p) < 0)
    CH.pop_back();
  CH.push_back(p);
}
set<point> sol;
void EXEC() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    int S, R;
    cin >> S >> R;
    a.push_back({S, R});
  }
  p = a;
  sort(p.begin(), p.end());
  for (int i = 0; i < p.size(); ++i)
    if (!i || p[i].x != p[i - 1].x) {
      Push(p[i]);
    }
  while (CH.size() >= 2 && CH.rbegin()->y <= (++CH.rbegin())->y) CH.pop_back();
  for (auto i : CH) sol.insert(i);
  for (int i = 0; i < n; ++i)
    if (sol.count(a[i])) printf("%d ", i + 1);
  printf("\n");
}
int main() {
  EXEC();
  return 0;
}
