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;
struct people {
  long long a, b, id;
} p[300010];
long long n, s[300010], t;
bool v[300010], ans[300010];
bool cmp(people a, people b) { return a.a == b.a ? a.b > b.b : a.a > b.a; }
long double check(people a, people b) {
  return (long double)a.a * b.a * (b.b - a.b) / (b.a - a.a) / b.b / a.b;
}
int main() {
  scanf("%lld", &n);
  for (int i = 1; i <= n; i++) scanf("%lld%lld", &p[i].a, &p[i].b), p[i].id = i;
  sort(p + 1, p + n + 1, cmp);
  s[1] = 1, t = 1;
  long long maxb = p[1].b;
  for (int i = 2; i <= n; i++)
    if (p[i].b <= maxb)
      v[i] = 1;
    else
      maxb = p[i].b;
  for (int i = 2; i <= n; i++) {
    if (v[i] || check(p[i], p[s[t]]) > 0) continue;
    while (t > 1 && check(p[i], p[s[t]]) < check(p[s[t - 1]], p[s[t]])) t--;
    s[++t] = i;
  }
  for (int i = 1; i <= t; i++) {
    ans[p[s[i]].id] = 1;
    for (int j = s[i] + 1; j <= n && p[s[i]].a == p[j].a && p[s[i]].b == p[j].b;
         j++)
      ans[p[j].id] = 1;
  }
  for (int i = 1; i <= n; i++)
    if (ans[i]) printf("%d ", i);
  return 0;
}
