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 Prg4
{
	public static void main(String args[])
	{
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt(),i,j;
		int a,b;
		ArrayList<Prac> ob1=new ArrayList<>();
		ArrayList<Prac> ob2=new ArrayList<>();
		for(i=0;i<n;i++){
			a=sc.nextInt();
			b=sc.nextInt();
			if(a<b)
				ob1.add(new Prac(a,b,i+1));
			else
				ob2.add(new Prac(a,b,i+1));
		}
		if(ob1.size()>=ob2.size()){
			Collections.sort(ob1, new ValueComp());
			System.out.println(ob1.size());
			for(i=ob1.size()-1;i>=0;i--){
				System.out.print(ob1.get(i).i+" ");
			}
		}
		else{
			Collections.sort(ob2, new ValueComp());
			System.out.println(ob2.size());
			for(i=0;i<ob2.size();i++){
				System.out.print(ob2.get(i).i+" ");
			}	
		}
		
	}
}

class Prac{
	int a;
	int b;
	int i;
	Prac(int a, int b, int i){
		this.a=a;
		this.b=b;
		this.i=i;
	}
}

class ValueComp implements Comparator<Prac>{  
	public int compare(Prac s1,Prac s2){  
		if(s1.a==s2.a)  
			return 0;  
		else if(s1.a>s2.a)  
			return 1;  
		else  
			return -1;  
	}  
}