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 int MAXN = 200005;
map<pair<long long int, long long int>, vector<int> > M;
vector<pair<long long int, long long int> > A;
pair<long long int, long long int> lower_left;
bool ans[MAXN];
bool sorter1(const pair<long long int, long long int> &x,
             const pair<long long int, long long int> &y) {
  return (x > y);
}
bool sorter2(const pair<long long int, long long int> &x,
             const pair<long long int, long long int> &y) {
  long long int num1 = (lower_left.first - x.first) * x.second * y.first *
                       (lower_left.second - y.second);
  long long int num2 = x.first * (lower_left.second - x.second) *
                       (lower_left.first - y.first) * y.second;
  if (num1 != num2)
    return num1 > num2;
  else
    return (x > y);
}
vector<pair<long long int, long long int> > find_hull(
    vector<pair<long long int, long long int> > &A) {
  sort(A.begin(), A.end(), sorter1);
  lower_left = A[0];
  sort(A.begin() + 1, A.end(), sorter2);
  if (A.size() < 3) return A;
  vector<pair<long long int, long long int> > ans(A.begin(), A.begin() + 2);
  int sz = 2;
  for (int i = 2; i < A.size(); ++i) {
    lower_left = ans[sz - 2];
    while (sz >= 2 && sorter2(A[i], ans[sz - 1])) {
      ans.pop_back();
      sz--;
      if (sz >= 2) lower_left = ans[sz - 2];
    }
    ans.push_back(A[i]);
    sz++;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  pair<long long int, long long int> temp;
  for (int i = 0; i < n; ++i) {
    cin >> temp.first >> temp.second;
    M[temp].push_back(i);
  }
  for (map<pair<long long int, long long int>, vector<int> >::iterator it =
           M.begin();
       it != M.end(); ++it)
    A.push_back(it->first);
  A = find_hull(A);
  lower_left = A[0];
  for (int i = 0; i < M[lower_left].size(); ++i) {
    ans[M[lower_left][i]] = true;
  }
  for (int i = 1; i < A.size(); ++i) {
    if (A[i].second > A[i - 1].second) {
      for (int j = 0; j < M[A[i]].size(); ++j) {
        ans[M[A[i]][j]] = true;
      }
    } else
      break;
  }
  for (int i = 0; i < n; ++i) {
    if (ans[i]) cout << i + 1 << " ";
  }
  return 0;
}
