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.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.Scanner;

public class Main {
	
	public static void main(String[] args)
	{
		Scanner s=new Scanner(System.in);
		
		int n=s.nextInt();
		
		ArrayList<pair> less=new ArrayList<>();
		ArrayList<pair> more=new ArrayList<>();
		
		for(int i=0;i<n;i++)
		{
			int a=s.nextInt();
			int b=s.nextInt();
			
			pair p=new pair(a,b,i+1);
			
			if(a<b)
			{
				less.add(p);
			}
			else
			{
				more.add(p);
			}
		}
		
		Collections.sort(less,new comp1());
		Collections.sort(more,new comp2());
		
		if(less.size()>=more.size())
		{
			System.out.println(less.size());
			
			for(int i=0;i<less.size();i++)
			{
				System.out.print(less.get(i).index+" ");
			}
		}
		else
		{
			System.out.println(more.size());
			
			for(int i=0;i<more.size();i++)
			{
				System.out.print(more.get(i).index+" ");
			}
		}
		
	}
	
}

class pair
{
	int a;
	int b;
	int index;
	
	public pair(int a,int b,int index)
	{
		this.a=a;
		this.b=b;
		this.index=index;
	}
}

class comp1 implements Comparator<pair>
{
	public int compare(pair a,pair b)
	{
		return b.b-a.b;
	}
}

class comp2 implements Comparator<pair>
{
	public int compare(pair a,pair b)
	{
		return a.a-b.a;
	}
}