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 node {
  long long x, y;
  int id;
  bool operator<(const node& b) const {
    if (y != b.y)
      return y < b.y;
    else
      return x < b.x;
  }
} a[210000], b[210000], c, d[210000];
vector<node> q;
int n, num, last;
int ans[210000];
set<node> s;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++)
    cin >> a[i].x >> a[i].y, a[i].id = i, d[i] = a[i];
  sort(a + 1, a + 1 + n);
  for (int i = n; i >= 1; i--)
    if (a[i].x > last) last = a[i].x, b[++num] = a[i];
  reverse(b + 1, b + 1 + num);
  for (int i = 1; i <= num; i++) {
    while (q.size() > 1) {
      node c = q[q.size() - 1], d = q[q.size() - 2];
      if ((c.y - d.y) * d.x * b[i].y * (c.x - b[i].x) <
          (b[i].y - c.y) * b[i].x * d.y * (d.x - c.x))
        q.pop_back();
      else
        break;
    }
    q.push_back(b[i]);
  }
  num = 0;
  for (int i = 0; i < q.size(); i++) s.insert(q[i]);
  for (int i = 1; i <= n; i++)
    if (s.count(d[i])) printf("%d ", i);
  return 0;
}
