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.util.*;

public class Main {

	static ArrayList<Integer> solveLess(ArrayList<pair> less) {
	
		ArrayList<Integer> ans=new ArrayList<Integer>();
		int lastR=Integer.MAX_VALUE;
		for(pair p:less) {
			if(p.a<lastR) {
				ans.add(p.idx);
				lastR=p.b;
			}
		}
		return ans;
	}
	static ArrayList<Integer> solveMore(ArrayList<pair> a) {
		Collections.sort(a,(x,y)->y.a-x.a);
		ArrayList<Integer> ans=new ArrayList<Integer>();

		int lastFirst=Integer.MAX_VALUE;
		for(pair p:a) {
			if(p.b<lastFirst) {
				ans.add(p.idx);
				lastFirst=p.a;
			}
		}
		return ans;
	}
	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner();
		PrintWriter out = new PrintWriter(System.out);
		int n=sc.nextInt();
		ArrayList<pair>less=new ArrayList(),more=new ArrayList();
		for(int i=0;i<n;i++) {
			int x=sc.nextInt(),y=sc.nextInt();
			pair p=new pair(x,y,i);
			if(x<y)
				less.add(p);
			else
				more.add(p);
		}
		Collections.sort(less,(x,y)->y.b-x.b);
		ArrayList<Integer> a=solveLess(less),b=solveMore(more);
		if(a.size()>b.size()) {
			out.println(a.size());
			for(int idx:a)
				out.print(idx+" ");
		}
		else {
			out.println(b.size());
			for(int i=b.size()-1;i>=0;i--) {
				out.print(b.get(i)+" ");
			}
		}
		
		out.close();

	}
	
	static class pair{
		int a,b,idx;
		pair(int x,int y,int i){
			a=x;
			b=y;
			idx=i+1;
		}
	}
	
	static class Scanner {
		BufferedReader br;
		StringTokenizer st;

		Scanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		Scanner(String fileName) throws FileNotFoundException {
			br = new BufferedReader(new FileReader(fileName));
		}

		String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		String nextLine() throws IOException {
			return br.readLine();
		}

		int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		long nextLong() throws NumberFormatException, IOException {
			return Long.parseLong(next());
		}

		double nextDouble() throws NumberFormatException, IOException {
			return Double.parseDouble(next());
		}

		boolean ready() throws IOException {
			return br.ready();
		}

	}

}