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;
long long arr[4][200009];
map<long long, long long> a, b;
int main() {
  long long n;
  cin >> n;
  for (long long f = 0; f < n; f++) {
    cin >> arr[0][f] >> arr[1][f];
    a[arr[0][f]] = max(a[arr[0][f]], arr[1][f]);
    b[arr[1][f]] = max(b[arr[1][f]], arr[0][f]);
  }
  vector<pair<long long, long long> > v;
  for (long long f = 0; f < n; f++) {
    v.push_back(make_pair(arr[0][f], arr[1][f] - arr[0][f]));
  }
  sort(v.begin(), v.end());
  long long mini = 1e9;
  for (long long f = 0; f < n; f++) {
    mini = min(mini, v[f].second);
    arr[2][f] = mini;
  }
  long long maxi = -1e9;
  for (long long f = n - 1; f >= 0; f--) {
    maxi = max(maxi, v[f].second);
    arr[3][f] = maxi;
  }
  for (long long f = 0; f < n; f++) {
    if (a[arr[0][f]] > arr[1][f] || b[arr[1][f]] > arr[0][f]) continue;
    long long i =
        lower_bound(v.begin(), v.end(), make_pair(arr[0][f], -1000000000ll)) -
        v.begin() - 1;
    long long j =
        upper_bound(v.begin(), v.end(), make_pair(arr[0][f], 1000000000ll)) -
        v.begin();
    if (j < v.size()) {
      if (arr[3][j] + arr[0][f] - arr[1][f] >= 0) continue;
    }
    if ((arr[0][f] - v[0].first) * (arr[3][j] + arr[0][f] - arr[1][f]) >
        (arr[0][f] - v[0].first) * (arr[2][j] + arr[0][f] - arr[1][f]))
      continue;
    cout << f + 1 << " ";
  }
  cout << endl;
}
