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 N = 200000 + 10;
int n;
int r[N], s[N];
pair<long long, long long> a[N];
pair<long long, long long> b[N];
void solve() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    r[i] = y;
    s[i] = x;
    a[i] = make_pair(y, x);
  }
  sort(a, a + n);
  int mv = -1;
  int m = 0;
  for (int i = n - 1; i >= 0; --i) {
    if (a[i].second <= mv) {
      continue;
    }
    mv = a[i].second;
    b[m++] = a[i];
  }
  reverse(b, b + m);
  vector<pair<long long, long long> > st;
  for (int i = 0; i < m; ++i) {
    pair<long long, long long> u, v, w;
    w = b[i];
    for (; st.size() > 1;) {
      v = st.back();
      u = st[st.size() - 2];
      if ((v.first - u.first) * u.second * w.first * (v.second - w.second) <
          (w.first - v.first) * w.second * u.first * (u.second - v.second)) {
        st.pop_back();
      } else {
        break;
      }
    }
    st.push_back(w);
  }
  set<pair<long long, long long> > ss;
  for (int i = 0; i < st.size(); ++i) {
    ss.insert(make_pair(st[i].first, st[i].second));
  }
  for (int i = 0; i < n; ++i) {
    if (ss.count(make_pair(r[i], s[i]))) {
      printf("%d ", i + 1);
    }
  }
}
int main() {
  solve();
  return 0;
}
