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.io.*;
import java.util.*;

public class Main{
    public static void main(String[] args)throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer zer = null;
        int n = Integer.parseInt(br.readLine());
        Pair[] a = new Pair[n];
        int t1,t2;
        for(int i = 0; i < n; i++){
            zer = new StringTokenizer(br.readLine());
            t1 = Integer.parseInt(zer.nextToken());
            t2 = Integer.parseInt(zer.nextToken());
            a[i] = new Pair(t1,t2,i+1);
        } 
        ArrayList<Pair> g,l;
        g = new ArrayList<>();
        l = new ArrayList<>();
        for(int i = 0; i < n; i++){
            if(a[i].first < a[i].second){
                l.add(a[i]);
            }else{
                g.add(a[i]);
            }
        }
        System.out.println(Math.max(l.size(),g.size()));
        if(g.size() > l.size()){
            Collections.sort(g,new SortHigher());
            for(Pair i:g) System.out.print(i.index+" ");
        }else{
            Collections.sort(l,new SortLower());
            for(Pair i:l) System.out.print(i.index+" ");
        }
    }    
}

class Pair{
    int first,second;
    int index;
    
    public Pair(int _f, int _s, int _i){
        this.first = _f;
        this.second = _s;
        index = _i;
    }
}

class SortHigher implements Comparator<Pair>{
    public int compare(Pair a,Pair b){
        return a.first - b.first;
    }
}

class SortLower implements Comparator<Pair>{
    public int compare(Pair a,Pair b){
        return b.second - a.second;
    }
}