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 
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>
#pragma GCC optimize("O2,unroll-loops")
using namespace std;
const int inf = 1000000010;
const long long INF = 1000000000000001000LL;
const int mod = 1000000007;
const int MAXN = 300010, LOG = 20;
const long double eps = 1e-9;
int n, m, k, u, v, x, y, t, a, b, sz;
pair<long double, long double> A[MAXN], B[MAXN];
pair<long double, pair<long double, long double> > cht[MAXN];
long double Inter(pair<long double, long double> a,
                  pair<long double, long double> b) {
  if (abs(a.first - b.first) < eps)
    return (a.second + eps < b.second ? -inf : inf);
  return (a.second - b.second) / (b.first - a.first);
}
void Add(pair<long double, long double> x) {
  while (sz && Inter(cht[sz].second, x) + eps < cht[sz].first) sz--;
  if (!sz)
    cht[++sz] = {-inf, x};
  else
    cht[sz + 1] = {Inter(cht[sz - 1].second, x), x}, sz++;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    long double x, y;
    cin >> x >> y;
    A[i] = B[i] = {1 / x, 1 / y};
  }
  sort(B + 1, B + n + 1);
  for (int i = 1; i <= n; i++) Add(B[i]);
  cerr << "sz" << '=' << (sz) << endl;
  for (int i = 1; i <= sz; i++) B[i] = cht[i].second;
  for (int i = 1; i <= n; i++) {
    int j = lower_bound(B + 1, B + sz + 1, A[i]) - B;
    if (j <= sz && B[j] == A[i]) cout << i << " ";
  }
  cout << "\n";
  return 0;
}
