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.lang.*;
import java.util.*;

public class D1148 {
    public static void main(String[] args) throws IOException{
        StringBuffer ans = new StringBuffer();
        StringTokenizer st;
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        st = new StringTokenizer(f.readLine());
        int n = Integer.parseInt(st.nextToken());
        ArrayList<point> arr1 = new ArrayList<>();
        ArrayList<point> arr2 = new ArrayList<>();
        for(int i = 0; i < n; i++){
            st = new StringTokenizer(f.readLine());
            int ft = Integer.parseInt(st.nextToken());
            int sc = Integer.parseInt(st.nextToken());
            if(ft > sc) {
                arr1.add(new point(ft, sc));
                arr1.get(arr1.size()-1).index = i+1;
            }
            else {
                arr2.add(new point(sc, ft));
                arr2.get(arr2.size()-1).index = i+1;
            }
        }
        Collections.sort(arr1);
        Collections.sort(arr2);
        if(arr1.size() > arr2.size()){
            ans.append(arr1.size());
            ans.append("\n");
            for(int i = arr1.size()-1; i > -1; i--){
                ans.append(arr1.get(i).index);
                ans.append(" ");
            }
        }else{
            ans.append(arr2.size());
            ans.append("\n");
            for(int i = 0; i < arr2.size(); i++){
                ans.append(arr2.get(i).index);
                ans.append(" ");
            }
        }
        f.close();


        System.out.println(ans);
    }
    public static class point implements Comparable<point>{
        int x;
        int y;
        int index = 0;
        public point(int x,int y){
            this.x = x;
            this.y = y;
        }
        public String toString(){
            return(x + " " + y);
        }
        public boolean equals(Object x){
            point y = ((point)(x));
            if (this.x == y.x && this.y == y.y){
                return true;
            }
            return false;
        }
        public int hashCode(){
            return Objects.hash(x,y);
        }
        public int compareTo(point other){
            if(this.x > other.x){
                return -1;
            }else if(this.x == other.x){
                return 0;
            }
            return 1;
        }
    }
}