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 
#include <bits/stdc++.h>
using namespace std;
char f(const pair<long long, long long> &a, const pair<long long, long long> &b,
       const pair<long long, long long> &c) {
  return (c.second - a.second) * (c.first - b.first) * a.first * b.second <
         (c.first - a.first) * (c.second - b.second) * a.second * b.first;
}
int main() {
  int n;
  scanf("%d", &n);
  vector<pair<long long, long long> > p(n);
  for (pair<long long, long long> &i : p) {
    scanf("%I64d%I64d", &i.first, &i.second);
  }
  vector<pair<long long, long long> > P = p, ch;
  p.emplace_back(0, 1e4 + 1);
  p.emplace_back(1e4 + 1, 0);
  sort((p).begin(), (p).end());
  for (auto i : p) {
    while (ch.size() > 1 && (ch.back().second == i.second ||
                             f(ch[(ch.size() - 2)], ch.back(), i))) {
      ch.pop_back();
    }
    ch.push_back(i);
  }
  for (int i = 0; i < n; ++i)
    if (binary_search((ch).begin(), (ch).end(), P[i])) {
      printf("%d ", i + 1);
    }
  return 0;
}
