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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
uniform_int_distribution<int> gen(1, 10);
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long fac(long long n) { return (n > 1) ? n * fac(n - 1) : 1; }
int dgts(int i) { return i > 0 ? (int)log10((double)i) + 1 : 1; }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<tuple<int, int, int>> mode1;
  vector<tuple<int, int, int>> mode2;
  for (int i = 0; i < n; i++) {
    int a, b;
    cin >> a >> b;
    if (a > b)
      mode1.push_back(make_tuple(a, b, i + 1));
    else
      mode2.push_back(make_tuple(a, b, i + 1));
  }
  sort(mode1.begin(), mode1.end());
  sort(mode2.rbegin(), mode2.rend());
  if (mode1.size() > mode2.size()) {
    cout << mode1.size() << '\n';
    for (auto e : mode1) {
      cout << get<2>(e) << " ";
    }
  } else {
    cout << mode2.size() << '\n';
    for (auto e : mode2) {
      cout << get<2>(e) << " ";
    }
  }
  return 0;
}