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;
template <class T>
T sqr(T first) {
  return first * first;
}
const int maxn = 200011;
vector<pair<long long, long long> > a, origin, ans_list;
int s[maxn];
int n, top;
bool gao(pair<long long, long long> &a, pair<long long, long long> &b,
         pair<long long, long long> &c) {
  return (c.second - a.second) * (c.first - b.first) * (a.first * b.second) <
         (c.first - a.first) * (c.second - b.second) * (a.second * b.first);
}
int main() {
  cin >> n;
  pair<long long, long long> tt;
  for (int i = 1; i <= n; i++) {
    cin >> tt.first >> tt.second;
    a.push_back(tt);
    origin.push_back(tt);
  }
  a.push_back(pair<long long, long long>(1e5 + 1, 0));
  a.push_back(pair<long long, long long>(0, 1e5 + 1));
  sort(a.begin(), a.end());
  for (int i = 0; i <= n + 1; i++) {
    while (top > 1 && (a[s[top]].second == a[i].second ||
                       gao(a[s[top - 1]], a[s[top]], a[i])))
      top--;
    s[++top] = i;
  }
  for (int i = 1; i <= top; i++) {
    ans_list.push_back(a[s[i]]);
  }
  sort(ans_list.begin(), ans_list.end());
  for (int i = 0; i <= n - 1; i++)
    if (binary_search(ans_list.begin(), ans_list.end(), origin[i]))
      printf("%d ", i + 1);
  return 0;
  fclose(stdin);
}
