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 JAVA):

Solution:
import java.util.*;


public class CP {
    public static void main(String[] args) {
        List<Pair> type2 = new ArrayList<>(), type1 = new ArrayList<>();
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        for(int i = 1; i <= n; i++) {
            int a = scanner.nextInt(), b = scanner.nextInt();
            if(a<b)type1.add(new Pair(a, b, i));
            else type2.add(new Pair(a, b, i));
        }
        int s1 = type1.size();
        int s2 = type2.size();

        if(s1>=s2) {
            System.out.println(s1);
            type1.sort(new ComparatorType1());
            for(Pair p: type1) {
                System.out.print(p.idx + " ");
            }
        } else {
            System.out.println(s2);
            type2.sort(new ComparatorType2());
            for(Pair p: type2) {
                System.out.print(p.idx + " ");
            }
        }
    }
}

class Pair {
    public int a, b, idx;

    public Pair(int a, int b, int idx) {
        this.a = a;
        this.b = b;
        this.idx = idx;
    }
}

class ComparatorType1 implements Comparator<Pair>{
    @Override
    public int compare(Pair o1, Pair o2) {
        return o1.a > o2.a?-1:1;
    }
}

class ComparatorType2 implements Comparator<Pair> {
    @Override
    public int compare(Pair o1, Pair o2) {
        return o1.b < o2.b? -1: 1;
    }
}