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.util.Scanner;

public class codeForcesGlobalRound3D {
	
	static boolean ascendingFlag=true;
	public static void main(String args[]) {
		Scanner sc=new Scanner(System.in);
		
		int n=sc.nextInt();
		long[][] mainArray=new long[n][3];
		
		int ascendingPairs=0;
		for(int i=0;i<n;i++) {
			mainArray[i][0]=sc.nextLong();
			mainArray[i][1]=sc.nextLong();
			mainArray[i][2]=i+1;
			if(mainArray[i][0]<mainArray[i][1]) {
				ascendingPairs++;
			}
		}
		
		if(ascendingPairs<n-ascendingPairs) {
			ascendingFlag=false;
		}
		
		java.util.Arrays.sort(mainArray, new java.util.Comparator<long[]>() {
		    public int compare(long[] a, long[] b) {
		    	if(ascendingFlag==true)
		    		return Long.compare(b[0], a[0]);
		    	else
		    		return Long.compare(a[0], b[0]);
		    }
		});
		
		/*
		 * for(int i=0;i<n;i++) {
		 * System.out.println(mainArray[i][0]+" "+mainArray[i][1]); }
		 */
		
		if(ascendingFlag==true) {
			System.out.println(ascendingPairs);
			for(int i=0;i<n;i++) {
				if(mainArray[i][0]<mainArray[i][1]) {
					System.out.print(mainArray[i][2]+" ");
				}
			}
		}
		else {
			System.out.println(n-ascendingPairs);
			for(int i=0;i<n;i++) {
				if(mainArray[i][0]>mainArray[i][1]) {
					System.out.print(mainArray[i][2]+" ");
				}
			}
		}
		
		sc.close();
	}
	
}