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.util.*;
import java.util.Collections;
import java.util.ArrayList;
public class mohamedy {

    public static void main(String[] args) {
        // write your code here
        //String str1 = "converse";
        //String str2 = "conserve";


        //System.out.println(areMetaStrings(str1, str2) ? "Yes" : "No");
        //PrintWriter pw=new PrintWriter(System.out,true);
        Scanner sc = new Scanner(System.in);
       // Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        ArrayList<Pair>a1=new ArrayList<Pair>();
        ArrayList<Pair>a2=new ArrayList<Pair>();
        for(int i=0;i<n;i++){
            int a=sc.nextInt(),b=sc.nextInt();
            if(a<b)a1.add(new Pair(a,b,i+1));
            else a2.add(new Pair(a,b,i+1));
        }
        if(a1.size()>a2.size()){
            System.out.println(a1.size());
            a1.sort((x, y) -> y.first - x.first);
            for(Pair p:a1 )System.out.print(p.idx+" ");
        }else{
            System.out.println(a2.size());
            a2.sort((x, y) -> x.first - y.first);
            for(Pair p:a2)System.out.print(p.idx+" ");
        }



    }
    public static String sortString(String s){
        char c[]=s.toCharArray();
        Arrays.sort(c);
        return new String(c);
    }

    static class Pair  {
        int idx;
        int first, second;

        Pair(int a, int b, int c) {
            first = a;
            second = b;
            idx = c;
        }

    }
/*
    static class Pair implements Comparable<Pair> {
        int a, b, idx;

        public Pair(int a, int b, int idx) {
            this.a = a;
            this.b = b;
            this.idx = idx;
        }

        @Override
        public int compareTo(Pair o) {
            return a - o.a;
        }
    }
*/
}