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;
set<pair<long long, long long> > s;
pair<long long, long long> a[300000], b[300000];
int num = 0, n, p[300000], A[300000], B[300000], cnt = 0;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++)
    cin >> A[i] >> B[i], a[i] = pair<long long, long long>(A[i], B[i]);
  sort(a + 1, a + n + 1);
  int Min = -1;
  for (int i = n; i >= 1; i--)
    if (a[i].second > Min) Min = a[i].second, b[++num] = a[i];
  reverse(b + 1, b + num + 1);
  for (int i = 1; i <= num; i++) {
    pair<long long, long long> u = b[i];
    while (cnt > 1 &&
           ((b[p[cnt]].first - b[p[cnt - 1]].first) *
                (b[p[cnt]].second - u.second) * u.first * b[p[cnt - 1]].second <
            (b[p[cnt - 1]].second - b[p[cnt]].second) *
                (u.first - b[p[cnt]].first) * b[p[cnt - 1]].first * u.second))
      cnt--;
    p[++cnt] = i;
  }
  for (int i = 1; i <= cnt; i++) s.insert(b[p[i]]);
  for (int i = 1; i <= n; i++)
    if (s.count(pair<long long, long long>(A[i], B[i]))) cout << i << ' ';
  cout << '\n';
  return 0;
}
