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;
const int Nmax = 200005, Vmax = 10005;
pair<pair<int, int>, int> A[Nmax];
bool cannotWin[Nmax];
vector<pair<int, int>> firstC[Vmax], secondC[Vmax];
int main() {
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  for (int i = 1; i <= N; ++i) {
    cin >> A[i].first.first >> A[i].first.second;
    A[i].second = i;
    firstC[A[i].first.first].push_back({A[i].first.second, i});
    secondC[A[i].first.second].push_back({A[i].first.first, i});
  }
  for (int i = 1; i < Vmax; ++i) {
    int maxs = 0;
    for (auto& p : firstC[i]) maxs = max(maxs, p.first);
    for (auto& p : firstC[i])
      if (p.first < maxs) cannotWin[p.second] = true;
    maxs = 0;
    for (auto& p : secondC[i]) maxs = max(maxs, p.first);
    for (auto& p : secondC[i])
      if (p.first < maxs) cannotWin[p.second] = true;
  }
  int newN = 0;
  for (int i = 1; i <= N; ++i)
    if (!cannotWin[i]) A[++newN] = A[i];
  N = newN;
  set<pair<int, int>> S;
  sort(A + 1, A + N + 1);
  for (int i = 1, j; i <= N; i = j) {
    for (j = i; j <= N && A[j].first == A[i].first; ++j)
      ;
    auto it = S.lower_bound(make_pair(A[i].first.second, 0));
    S.erase(S.begin(), it);
    for (int k = i; k < j; ++k)
      S.insert(make_pair(A[k].first.second, A[k].second));
  }
  vector<int> ans;
  for (auto& p : S) ans.push_back(p.second);
  sort(ans.begin(), ans.end());
  for (int p : ans) cout << p << ' ';
  cout << endl;
}
