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;
const long double eps = 1e-25, inf = 2e+15;
pair<int, int> v[200010], v1[200010];
map<pair<int, int>, char> sol;
int v2[200010];
int cmp(pair<int, int> a, pair<int, int> b) {
  if (a.first == b.first)
    return a.second > b.second;
  else
    return a.first < b.first;
}
int scoate(pair<int, int> a, pair<int, int> b, pair<int, int> c) {
  return (
      1LL * a.first * c.second * (a.second - b.second) * (c.first - b.first) >
      1LL * a.second * c.first * (b.first - a.first) * (b.second - c.second));
}
int main() {
  int n, nr = 0, nr2 = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &v[i].first, &v[i].second);
    v1[i] = v[i];
  }
  sort(v1 + 1, v1 + 1 + n, cmp);
  for (int i = 1; i <= n; i++)
    if (v1[i].first != v1[nr].first) v1[++nr] = v1[i];
  for (int i = 1; i <= nr; i++) {
    while (nr2 >= 2 && scoate(v1[v2[nr2 - 1]], v1[v2[nr2]], v1[i])) nr2--;
    v2[++nr2] = i;
  }
  sol[v1[v2[nr2]]] = 1;
  for (int i = 1; i < nr2; i++)
    if (v1[v2[i]].second > v1[v2[i + 1]].second) sol[v1[v2[i]]] = 1;
  for (int i = 1; i <= n; i++)
    if (sol[v[i]]) printf("%d ", i);
  return 0;
}
