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;
const int N = 3e5 + 5;
struct node {
  int id, a, b;
  node() {}
  node(int _id, int _a, int _b) : id(_id), a(_a), b(_b) {}
};
vector<node> L, R;
bool comp1(node x, node y) { return x.b > y.b; }
bool comp2(node x, node y) { return x.b < y.b; }
map<pair<int, int>, bool> me;
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int a, b;
    cin >> a >> b;
    if (a > b && !me[make_pair(a, b)])
      L.push_back(node(i, a, b));
    else if (a < b && !me[make_pair(a, b)])
      R.push_back(node(i, a, b));
    me[make_pair(a, b)] = true;
  }
  sort(L.begin(), L.end(), comp2);
  sort(R.begin(), R.end(), comp1);
  if (L.size() > R.size()) {
    cout << L.size() << endl;
    for (auto it : L) cout << it.id << " ";
    cout << endl;
  } else {
    cout << R.size() << endl;
    for (auto it : R) cout << it.id << " ";
    cout << endl;
  }
}