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 int mxN = 2e5;
int n, qt;
long long a[mxN], b[mxN];
vector<int> c[mxN], ans;
struct tr {
  int first, second, i;
  inline bool operator<(const tr &o) const {
    return first == o.first ? second > o.second : first < o.first;
  }
} trs[mxN];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i)
    cin >> trs[i].first >> trs[i].second, trs[i].i = i;
  sort(trs, trs + n);
  for (int i = 0; i < n; ++i) {
    if (i && trs[i].first == a[qt - 1]) {
      if (trs[i].second == b[qt - 1]) c[qt - 1].push_back(trs[i].i);
      continue;
    }
    while (
        (qt >= 1 && trs[i].second >= b[qt - 1]) ||
        (qt >= 2 && a[qt - 2] * trs[i].second * (b[qt - 2] - b[qt - 1]) *
                            (trs[i].first - a[qt - 1]) >
                        trs[i].first * b[qt - 2] * (b[qt - 1] - trs[i].second) *
                            (a[qt - 1] - a[qt - 2])))
      c[--qt].clear();
    a[qt] = trs[i].first, b[qt] = trs[i].second;
    c[qt].push_back(trs[i].i);
    ++qt;
  }
  for (int i = 0; i < qt; ++i)
    for (int ci : c[i]) ans.push_back(ci);
  sort(ans.begin(), ans.end());
  for (int ai : ans) cout << ai + 1 << " ";
}
