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.ArrayList;
import java.util.Collections;
import java.util.Scanner;

public class Main {
	public static int N;
	public static ArrayList<segment> list1, list2;

	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		N = s.nextInt();
		list1 = new ArrayList<segment>();
		list2 = new ArrayList<segment>();
		for (int i = 0; i < N; i++) {
			int a = s.nextInt();
			int b = s.nextInt();
			if (a < b) {
				list1.add(new segment(a, b, i + 1));
			} else {
				list2.add(new segment(a, b, i + 1));
			}
		}
		if (list1.size() > list2.size()) {
			int M = list1.size();
			StringBuilder sb = new StringBuilder();
			Collections.sort(list1);
			for (int i = 0; i < M; i++) {
				sb.append(list1.get(i).index).append(" ");
			}
			System.out.println(M);
			System.out.print(sb.toString());
		} else {
			int M = list2.size();
			StringBuilder sb = new StringBuilder();
			Collections.sort(list2);
			Collections.reverse(list2);
			for (int i = 0; i < M; i++) {
				sb.append(list2.get(i).index).append(" ");
			}
			System.out.println(M);
			System.out.print(sb.toString());
		}
	}

	public static void printArray(int[] array) {
		for (int i : array) {
			System.out.print(i + " ");
		}
		System.out.println();
	}

	public static void printMatrix(int[][] matrix) {
		for (int[] is : matrix) {
			printArray(is);
		}
		System.out.println();
	}

	public static void printArray(long[] array) {
		for (long i : array) {
			System.out.print(i + " ");
		}
		System.out.println();
	}

	public static void printMatrix(long[][] matrix) {
		for (long[] is : matrix) {
			printArray(is);
		}
		System.out.println();
	}
}

class segment implements Comparable<segment> {
	public int a, b, index;

	public segment(int c, int d, int i) {
		a = Math.min(c, d);
		b = Math.max(c, d);
		index = i;
	}

	public String toString() {
		return a + " " + b + " " + index;
	}

	public int compareTo(segment next) {
		return Integer.compare(next.a, a);
	}
}