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.lang.*;
import java.io.*;
 
/* Name of the class has to be "Main" only if the class is public. */
public class Codechef{
	public static void main (String[] args)throws IOException {
    try{
	   //BufferedReader br =new BufferedReader(new InputStreamReader(System.in));
	   Scanner sc = new Scanner(System.in);
	//int t=Integer.parseInt(br.readLine());
	int t=1;
	 //int t=sc.nextInt();
	   while(t-->0){
	 //int n=Integer.parseInt(br.readLine())
      int n = sc.nextInt();
      List<pair> one = new ArrayList<pair>();
      List<pair> second = new ArrayList<pair>();
      int a1,a2,id=0;
      int form1=0;
      int form2=0;
      for(int i=0;i<n;i++){
          a1=sc.nextInt();
          a2=sc.nextInt();
          id=i+1;
          if(a1<a2) one.add(new pair(a1,a2,id));
          else  second.add(new pair(a1,a2,id));
         
      }
    
          Collections.sort(one, new comp1());
           Collections.sort(second, new comp2());
           
           
           if(one.size()>second.size()){
          System.out.println(one.size());
         for(int i=0;i<one.size();i++){
             System.out.print(one.get(i).ind+" ");
         }
           }
     else{
          System.out.println(second.size());
         for(int i=0;i<second.size();i++){
             System.out.print(second.get(i).ind+" ");
         } 
     }
	//long d=Long.parseLong(br.readLine());
	//String[] s= br.readLine().trim().split(" ");
	// Map<Integer,Integer> m = new TreeMap<Integer,Integer>();
    //List<Character> f = new ArrayList<Character>();
    //List<Integer> s = new ArrayList<Integer>();
     //<Integer> v= new Vector<Integer>();
	 //Set<Integer> s= new HashSet<Integer>();
	 
	 

}	
}
catch(Exception e){
    
}
	}
 
}
 class pair{
    int x,y,ind;
    pair(int n,int m,int id){
        x=n;
        y=m;
        ind=id;
    }
}
class comp1 implements Comparator<pair>{
    public int compare(pair p1,pair p2){
        if(p1.x>p2.x)return -1;
        else return 1;
    }
}

class comp2 implements Comparator<pair>{
    public int compare(pair p1,pair p2){
        if(p1.y<p2.y)return -1;
        else return 1;
    }
}