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 Main {
	public static void main(String args[]) {new Main().run();}
	
	FastReader in = new FastReader();
	PrintWriter out = new PrintWriter(System.out);
	void run(){
	    work();
	    out.flush();
	}
	long mod=998244353L;
	int ret=99999999;
	void work() {
	    int n=in.nextInt();
	    ArrayList<int[]> l1=new ArrayList<>();
	    ArrayList<int[]> l2=new ArrayList<>();
	    for(int i=0;i<n;i++){
	        int s=in.nextInt();
	        int e=in.nextInt();
	        if(s>e){
	            l1.add(new int[]{s,i+1});
	        }else{
	            l2.add(new int[]{e,i+1});
	        }
	    }
	    if(l1.size()>l2.size()){
	        out.println(l1.size());
	        Collections.sort(l1,new Comparator<int[]>(){
	            public int compare(int[] arr1,int[] arr2){
	                return arr1[0]-arr2[0];
	            }
	        });
	        for(int[] a:l1){
	            out.print(a[1]+" ");
	        }
	    }else{
	        out.println(l2.size());
	        Collections.sort(l2,new Comparator<int[]>(){
	            public int compare(int[] arr1,int[] arr2){
	                return arr2[0]-arr1[0];
	            }
	        });
	        for(int[] a:l2){
	            out.print(a[1]+" ");
	        }
	    }
	    
	}
	
}
class FastReader
{
	BufferedReader br;
	StringTokenizer st;
 
	public FastReader()
	{
	    br=new BufferedReader(new InputStreamReader(System.in));
	}
 
	public String next() 
	{
	    if(st==null || !st.hasMoreElements())
	    {
	        try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
	    }
	    return st.nextToken();
	}
 
	public int nextInt() 
	{
	    return Integer.parseInt(next());
	}
 
	public long nextLong()
	{
	    return Long.parseLong(next());
	}
}