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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;
import java.util.StringTokenizer;

public class Submit {
    public static void main(String args[]) throws IOException {
        process();
    }
    public static void process() throws IOException{
        BufferedReader in=new BufferedReader(new InputStreamReader(System.in));
        int n=Integer.parseInt(in.readLine());
        int arr[][]=new int[n][2];
        Map<Integer,Integer> search=new HashMap<>();
        int a[]=new int[n];
        int b[]=new int[n];
        for(int i=0;i<n;i++) {
            a[i]=-1;
            b[i]=-1;
        }
        int a_count=0;
        int b_count=0;
        StringTokenizer st;
        for(int i=0;i<n;i++){
            st=new StringTokenizer(in.readLine()," ");
            arr[i][0]=Integer.parseInt(st.nextToken());
            arr[i][1]=Integer.parseInt(st.nextToken());
            search.put(arr[i][0],i+1);
            if(arr[i][0]>arr[i][1]){
                a[i]=arr[i][0];
                a_count++;
            }
            else{
                b[i]=arr[i][0];
                b_count++;
            }
        }
        if(a_count>b_count) {
            System.out.println(a_count);
            Arrays.sort(a);
            for (int i = 0; i < n; i++)
                if (a[i] != -1)
                    System.out.print(search.get(a[i])+" ");
        }
        else {
            System.out.println(b_count);
            Arrays.sort(b);
            for(int i=n-1;i>=0;i--)
                if (b[i] != -1)
                    System.out.print(search.get(b[i])+" ");
        }
    }
}