Tavas is a cheerleader in the new sports competition named "Pashmaks".

<image>

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 
5
10 50
10 50
10 50
10 50
10 50
 2
10000 10000
10000 10000
 3
10000 10000
10000 10000
10000 10000
 3
1000 3000
1500 1500
3000 1000
 43
75 16
70 7
75 16
54 2
70 7
54 2
70 7
54 2
70 7
54 2
54 2
70 7
70 7
54 2
70 7
75 16
70 7
54 2
54 2
54 2
54 2
54 2
54 2
75 16
70 7
70 7
54 2
70 7
75 16
54 2
75 16
75 16
54 2
54 2
70 7
70 7
75 16
70 7
70 7
70 7
75 16
75 16
54 2
 1
1 1
 2
1 1
1 1
 44
78 52
75 16
70 7
75 16
54 2
70 7
54 2
70 7
54 2
70 7
54 2
54 2
70 7
70 7
54 2
70 7
75 16
70 7
54 2
54 2
54 2
54 2
54 2
54 2
75 16
70 7
70 7
54 2
70 7
75 16
54 2
75 16
75 16
54 2
54 2
70 7
70 7
75 16
70 7
70 7
70 7
75 16
75 16
54 2
 3
1000 3000
1499 1499
3000 1000
 5
50 50
49 50
50 49
49 49
50 1
 18
82 38
33 69
33 69
33 69
33 69
33 69
82 38
33 69
33 69
33 69
82 38
33 69
33 69
82 38
82 38
33 69
33 69
33 69
 1
10000 9999
 3
1000 3000
1501 1501
3000 1000
 2
1 2
2 1

1 2 3 4 5
 1 2
 1 2 3
 1 2 3
 1 3 16 24 29 31 32 37 41 42
 1
 1 2
 1
 1 3
 1
 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
 1
 1 2 3
 1 2 

#include <bits/stdc++.h>
using namespace std;
const long long maxn = 2e5 + 100;
const long long mod = 1e9 + 7;
const long long eps = 1e-5;
const long long inf = 1e18;
long long n, x[maxn], y[maxn];
pair<pair<long long, long long>, long long> a[maxn];
vector<long long> vec;
bool ok;
long double get(long long i, long long j) {
  return 1.0 * (y[i] - y[j]) / (x[j] - x[i]);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  if (n == 122304) ok = 0;
  for (long long i = 0; i < n; i++) {
    cin >> a[i].first.first >> a[i].first.second;
    a[i].first.second *= -1;
    a[i].second = i + 1;
  }
  sort(a, a + n);
  for (long long i = 0; i < n; i++) {
    x[i] = a[i].first.first;
    y[i] = -a[i].first.second;
  }
  vec.push_back(0);
  for (long long i = 1; i < n; i++) {
    if (x[i] == x[i - 1] && y[i] != y[i - 1]) {
      if (ok && a[i].second == 52750) cout << "oh2\n";
      continue;
    }
    while (long long((vec).size()) > 1) {
      if (get(i, vec.back()) -
              get(vec.back(), vec[long long((vec).size()) - 2]) <
          eps)
        vec.pop_back();
      else
        break;
    }
    vec.push_back(i);
  }
  if (ok) {
    for (long long i = 0; i < n; i++) {
      if (a[i].second == 52750) cout << x[i] << " " << y[i] << '\n';
    }
  }
  vector<long long> ans;
  for (long long i = long long((vec).size()); i--;) {
    ans.push_back(a[vec[i]].second);
    if (ok) cout << x[vec[i]] << " " << y[vec[i]] << '\n';
    if (i > 0 && (y[vec[i]] >= y[vec[i - 1]] && x[vec[i]] != x[vec[i - 1]])) {
      if (ok) cout << x[vec[i - 1]] << " " << y[vec[i - 1]] << '\n';
      break;
    }
  }
  sort((ans).begin(), (ans).end());
  for (long long i : ans) {
    cout << i << " ";
  }
  return 0;
}
