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.Scanner;
import java.util.Vector;
import java.util.Comparator;

public class D{
	static class P{
		int a;
		int i;
		int b;
		P(int a, int b,int i){
			this.a = a;
			this.b = b;
			this.i = i;
		}
	}
	public static void main(String[] args){
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		Vector<P> aless =new Vector<P>();
		Vector<P> bless =new Vector<P>();
		for(int i=1;i<n+1;++i){
			P p = new P(in.nextInt(),in.nextInt(),i);
			if(p.a<p.b) aless.add(p);
			else bless.add(p);
		}
		Comparator<P> compb = new Comparator<P>(){
			public int compare(P p1, P p2){
				return Integer.compare(p1.b,p2.b);
			}};	
		Comparator<P> compa = new Comparator<P>(){
			public int compare(P p1, P p2){
				return -Integer.compare(p1.a,p2.a);
			}};		
		aless.sort(compa);
		bless.sort(compb);
		long res = Math.max(aless.size(), bless.size());
		StringBuilder str= new StringBuilder();
		if(aless.size()>bless.size()){
			for(int i = 0;i<aless.size();++i)
				str.append(aless.get(i).i + " ");
		} else {
			for(int i = 0;i<bless.size();++i)
				str.append(bless.get(i).i + " ");
		}
		System.out.println(res);
		System.out.println(str);
	}
}