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>
using namespace std;
int N;
int S[303030], T[303030];
int ma[11000];
multiset<int> SS[10100];
vector<int> R;
double winR[11000];
void solve() {
  int i, j, k, l, r, x, y, s;
  cin >> N;
  for (i = 0; i < N; i++) {
    cin >> S[i] >> T[i];
    if (ma[S[i]] < T[i]) ma[S[i]] = T[i], SS[S[i]].clear();
    if (ma[S[i]] == T[i]) SS[S[i]].insert(i);
  }
  y = -1;
  for (s = 10000; s >= 1; s--)
    if (ma[s] > 0) {
      if (y == -1)
        y = s, winR[s] = 1e-15;
      else if (ma[s] > ma[y]) {
        winR[s] = (1.0 / s - 1.0 / y) / ((1.0 / ma[y] - 1.0 / ma[s]));
      }
    }
  double mr = 1e20;
  for (s = 1; s <= 10000; s++)
    if (winR[s] > 0 && winR[s] < mr) {
      mr = winR[s];
      for (auto& r : SS[s]) R.push_back(r + 1);
    }
  sort(R.begin(), R.end());
  for (auto& r : R) (void)printf("%d ", r);
  (void)printf("\n");
}
int main(int argc, char** argv) {
  string s;
  int i;
  if (argc == 1) ios::sync_with_stdio(false);
  for (i = 0; i < argc - 1; i++) s += argv[i + 1], s += '\n';
  for (i = 0; i < s.size(); i++) ungetc(s[s.size() - 1 - i], stdin);
  solve();
  return 0;
}
