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:
//package com.company;
import java.io.*;
import java.util.*;
import java.math.*;
public class Main{
   static  Scanner s = new Scanner(System.in);
   static  StringBuilder sb= new StringBuilder();
    public static void main(String[] args)throws Exception {
    int n=s.nextInt();int th=0;int ia=0;int ib=0;int ia1=0;int ib1=0;
       int[] x1=new int[n];int[] x2=new int[n];
        for(int i=0;i<n;i++) {
            x1[i]=s.nextInt();x2[i]=s.nextInt();
         if(x2[i]-x1[i]>0) ia1++;
         else ib1++;
        }Student[] b=new Student[ib1];
        Student [] a=new Student[ia1];
            for(int i=0;i<n;i++){
                long x =x1[i];
                long y = x2[i];

                if(y-x>0){
            a[ia++]=new Student(x,y,i+1);
        }else b[ib++]=new Student(x,y,i+1);

    }
    Arrays.sort(a,new Sortbyroll());
    Arrays.sort(b,new Sortbyroll());StringBuilder sbb=new StringBuilder();StringBuilder sba=new StringBuilder();
    for(int i=0;i<ib;i++){
       sbb.append(b[i].i+" ");
    } for(int i=ia-1;i>=0;i--){
            sba.append(a[i].i+" ");
        }
        if(ia>ib){
            System.out.println(ia);
            System.out.println(sba.toString());
        }else{
            System.out.println(ib);
            System.out.println(sbb.toString());
        }
    }   public static int i(){
        return s.nextInt();
    }public static long l(){
        return s.nextLong();
    }public static long[] a(int n){
        long [] a=new long [n];
        for(int i=0;i<n;i++){
            a[i]=s.nextLong();
        }return a;
    }public static char[] c(){
        char[] a=s.next().toCharArray();
        return a;
    }
}class Student {
 long s,h;char[] string;int i;
    public Student(long a ,long b,int i) {
  this. s=a;this.h=b;this.i=i;//this.string=c.clone();
    }

}class Sortbyroll implements Comparator<Student>
{
    public int   compare(Student a, Student b)
    {   return (int)a.h-(int)b.h;}
}