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 std::cin;
using std::cout;
const long long N = 300007;
struct _ {
  long long a, b, idx;
} p[N];
bool cmp(_ a, _ b) {
  if (a.a == b.a) return a.b > b.b;
  return a.a > b.a;
}
double check(_ x, _ y) {
  return (double)x.a * y.a * (y.b - x.b) / (y.a - x.a) / y.b / x.b;
}
long long n, t, s[N], v[N], ans[N];
signed main() {
  std::ios::sync_with_stdio(0);
  cin >> n;
  for (long long i = 1; i <= n; ++i) {
    cin >> p[i].a >> p[i].b;
    p[i].idx = i;
  }
  std::sort(p + 1, p + 1 + n, cmp);
  s[1] = 1;
  t = 1;
  long long max = p[1].b;
  for (long long i = 2; i <= n; ++i) {
    if (p[i].b <= max)
      v[i] = 1;
    else
      max = p[i].b;
  }
  for (long long i = 2; i <= n; ++i) {
    if (v[i]) continue;
    if (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 (long long i = 1; i <= t; ++i) {
    ans[p[s[i]].idx] = 1;
    for (long long j = s[i] + 1;
         j <= n && p[s[i]].a == p[j].a && p[s[i]].b == p[j].b; ++j)
      ans[p[j].idx] = 1;
  }
  for (long long i = 1; i <= n; ++i)
    if (ans[i]) cout << i << ' ';
  return 0;
}
