Problem: You are given n pairs of integers (a_1, b_1), (a_2, b_2), …, (a_n, b_n). All of the integers in the pairs are distinct and are in the range from 1 to 2 ⋅ n inclusive.

Let's call a sequence of integers x_1, x_2, …, x_{2k} good if either 

  * x_1 < x_2 > x_3 < … < x_{2k-2} > x_{2k-1} < x_{2k}, or 
  * x_1 > x_2 < x_3 > … > x_{2k-2} < x_{2k-1} > x_{2k}. 



You need to choose a subset of distinct indices i_1, i_2, …, i_t and their order in a way that if you write down all numbers from the pairs in a single sequence (the sequence would be a_{i_1}, b_{i_1}, a_{i_2}, b_{i_2}, …, a_{i_t}, b_{i_t}), this sequence is good.

What is the largest subset of indices you can choose? You also need to construct the corresponding index sequence i_1, i_2, …, i_t.

Input

The first line contains single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the number of pairs.

Each of the next n lines contain two numbers — a_i and b_i (1 ≤ a_i, b_i ≤ 2 ⋅ n) — the elements of the pairs.

It is guaranteed that all integers in the pairs are distinct, that is, every integer from 1 to 2 ⋅ n is mentioned exactly once.

Output

In the first line print a single integer t — the number of pairs in the answer.

Then print t distinct integers i_1, i_2, …, i_t — the indexes of pairs in the corresponding order.

Examples

Input


5
1 7
6 4
2 10
9 8
3 5


Output


3
1 5 3


Input


3
5 4
3 2
6 1


Output


3
3 2 1

Note

The final sequence in the first example is 1 < 7 > 3 < 5 > 2 < 10.

The final sequence in the second example is 6 > 1 < 3 > 2 < 5 > 4.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n;
  cin >> n;
  map<pair<int, int>, int> m;
  std::vector<pair<int, int> > p1, p2;
  int x, y;
  for (int i = 0; i < n; ++i) {
    cin >> x >> y;
    m[make_pair(x, y)] = i + 1;
    if (y > x)
      p1.push_back(make_pair(x, y));
    else
      p2.push_back(make_pair(y, x));
  }
  if (p1.size() > p2.size()) {
    sort(p1.begin(), p1.end());
    cout << p1.size() << "\n";
    for (vector<pair<int, int> >::iterator i = p1.end() - 1; i >= p1.begin();
         --i)
      cout << m[*i] << " ";
  } else {
    sort(p2.begin(), p2.end());
    cout << p2.size() << "\n";
    for (auto i : p2) cout << m[make_pair(i.second, i.first)] << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long int t = 1;
  while (t--) {
    cerr << "'\\n'"
         << " " << '\n'
         << "\n";
    solve();
    cout << "\n";
  }
}