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.*;
import java.io.*;
public class Solution{

	static class Pair{
		int a, b, index;
		Pair(int a, int b, int index){
			this.a=a;
			this.b=b;
			this.index=index;
		}
	}

	public static void main(String args[])throws IOException{

		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));	Scanner in=new Scanner(System.in);	StringBuffer out=new StringBuffer();

		int n=INT(br.readLine());

		Pair[] pair=new Pair[n];
		for(int i=0; i<n; i++){
			String line[]=br.readLine().split("\\s");
			pair[i]=new Pair(INT(line[0]), INT(line[1]), i+1);
		}

		ArrayList<Pair> ab=new ArrayList<>(),
						ba=new ArrayList<>();

		for(Pair item: pair)
			if(item.a>item.b)
				ab.add(item);
			else
				ba.add(item);

		if(ab.size()>ba.size()){
			Collections.sort(ab, (Pair x, Pair y)->{
				return x.b-y.b;
			});
			for(Pair item: ab)
				out.append(item.index+" ");
		}
		else{
			Collections.sort(ba, (Pair x, Pair y)->{
				return y.a-x.a;
			});
			for(Pair item: ba)
				out.append(item.index+" ");
		}

		System.out.println(Math.max(ab.size(), ba.size()));
		System.out.println(out);
	}

	public static Integer INT(String s){
		return Integer.parseInt(s);
	}

	public static Long LONG(String s){
		return Long.parseLong(s);
	}
}