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;
vector<pair<pair<long long, long long>, int> > v, v1;
vector<pair<pair<long long, long long>, int> > cht;
map<long long, long long> my;
int c[200011] = {0};
vector<pair<long long, long long> > r;
multiset<long long> second;
multiset<long long>::iterator it;
bool intersection(int l1, int l2, int l3) {
  long long p1 = (cht[l2].first.second * cht[l3].first.first *
                  (cht[l3].first.second - cht[l1].first.second) *
                  (cht[l1].first.first - cht[l2].first.first));
  long long p2 = (cht[l3].first.second * cht[l2].first.first *
                  (cht[l2].first.second - cht[l1].first.second) *
                  (cht[l1].first.first - cht[l3].first.first));
  int p;
  if (p1 < p2)
    p = 1;
  else
    p = 0;
  return p;
}
void add(long long m, long long b, int i) {
  cht.push_back(make_pair(make_pair(m, b), i));
  while (cht.size() > 2 &&
         intersection(cht.size() - 3, cht.size() - 2, cht.size() - 1))
    cht.erase(cht.end() - 2);
}
int main() {
  int i, j, n, n1;
  cin >> n;
  n1 = n;
  for (i = 0; i < n; i++) {
    long long r1, s1;
    cin >> s1 >> r1;
    r.push_back(make_pair(s1, r1));
    v1.push_back(make_pair(make_pair(s1, r1), i + 1));
  }
  sort(v1.begin(), v1.end());
  for (i = 0; i < n; i++) second.insert(v1[i].first.second);
  for (i = 0; i < n; i++) {
    second.erase(second.find(v1[i].first.second));
    if (second.size() == 0) {
      v.push_back(make_pair(make_pair(v1[i].first.first, v1[i].first.second),
                            v1[i].second));
      continue;
    }
    it = second.lower_bound(v1[i].first.second);
    if (it == second.end())
      v.push_back(make_pair(make_pair(v1[i].first.first, v1[i].first.second),
                            v1[i].second));
  }
  n = v.size();
  for (i = 0; i < n; i++) add(v[i].first.first, v[i].first.second, v[i].second);
  for (i = 0; i < cht.size(); i++) {
    c[cht[i].second] = 1;
    my[cht[i].first.first] = cht[i].first.second;
  }
  for (i = 0; i < n1; i++)
    if (c[i + 1] == 0 && my.find(r[i].first) != my.end() &&
        my[r[i].first] == r[i].second)
      c[i + 1] = 1;
  for (i = 1; i <= n1; i++)
    if (c[i] == 1) printf("%d ", i);
  return 0;
}
