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.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.StringTokenizer;

public class D {
	
	public static void main(String[] args) throws IOException {
		MyScanner sc = new MyScanner();
		PrintWriter out = new PrintWriter(System.out);
		int n = sc.nextInt();
		List<Pair> up = new ArrayList<>();
		List<Pair> down = new ArrayList<>();
		for (int i = 0; i < n; i++) {
			int a = sc.nextInt();
			int b = sc.nextInt();
			Pair p = new Pair(i + 1, a, b);
			if (a < b) {
				up.add(p);
			} else {
				down.add(p);
			}
		}
		if (up.size() >= down.size()) {
			Collections.sort(up, (p1, p2) -> -Integer.compare(p1.a, p2.a));
			out.println(up.size());
			for (int i = 0; i < up.size(); i++) {
				if (i > 0) {
					out.print(" ");
				}
				out.print(up.get(i).id);
			}
		} else {
			Collections.sort(down, (p1, p2) -> Integer.compare(p1.a, p2.a));
			out.println(down.size());
			for (int i = 0; i < down.size(); i++) {
				if (i > 0) {
					out.print(" ");
				}
				out.print(down.get(i).id);
			}
		}
		out.println();
		out.flush();
	}
	
	static class Pair {
		int id, a, b;
		
		public Pair(int id, int a, int b) {
			this.id = id;
			this.a = a;
			this.b = b;
		}
		
		
	}
	
	static class MyScanner {
		private BufferedReader br;
		private StringTokenizer tokenizer;
		
		public MyScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		
		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}
		
		public int nextInt() {
			return Integer.parseInt(next());
		}
		
		public long nextLong() {
			return Long.parseLong(next());
		}
	}
}