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.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Stack;
import java.util.StringTokenizer;

public class Main {
	public static void main(String[] args) {
		Reader in = new Reader();
		int n = in.nextInt();
		ArrayList<Pair> inc = new ArrayList<>();
		ArrayList<Pair> dec = new ArrayList<>();
		for(int i = 0; i<n; i++) {
			Pair p = new Pair(in.nextInt(), in.nextInt(), i+1);
			if(p.inc) inc.add(p);
			else dec.add(p);
		}
		Collections.sort(inc);
		Collections.sort(dec);
		if(inc.size()>dec.size()) {
			System.out.println(inc.size());
			for(Pair p : inc) System.out.print(p.idx+" ");
		}else {
			System.out.println(dec.size());
			for(Pair p : dec) System.out.print(p.idx+" ");
		}
	}
	static class Pair implements Comparable<Pair>{
		int a,b,idx;
		boolean inc;
		public Pair(int a, int b,int idx) {
			this.a = a;
			this.b = b;
			this.idx = idx;
			inc = a<b;
		}
		@Override
		public int compareTo(Pair o) {
			if(inc) return o.a - this.a;
			else return this.b - o.b;
		}
	}
	static class Reader {
		static BufferedReader br;
		static StringTokenizer st;
		private int charIdx = 0;
		private String s;

		public Reader() {
			this.br = new BufferedReader(new InputStreamReader(System.in));
		}

		public int[] na(int n) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = nextInt();
			return a;
		}
		public double[] nd(int n) {
			double[] a = new double[n];
			for (int i = 0; i < n; i++)
				a[i] = nextDouble();
			return a;
		}
		public char nextChar() {
			if (s == null || charIdx >= s.length()) {
				if (st == null || !st.hasMoreTokens())
					try {
						readLine();
					} catch (Exception e) {
					}
				s = st.nextToken();
				charIdx = 0;
			}
			return s.charAt(charIdx++);
		}

		public long[] nl(int n) {
			long[] a = new long[n];
			for (int i = 0; i < n; i++)
				a[i] = nextLong();
			return a;
		}

		public char[] nca() {
			return next().toCharArray();
		}

		public String[] nS(int n) {
			String[] a = new String[n];
			for (int i = 0; i < n; i++)
				a[i] = next();
			return a;
		}

		public int nextInt() {
			if (st == null || !st.hasMoreTokens())
				try {
					readLine();
				} catch (Exception e) {
				}
			return Integer.parseInt(st.nextToken());
		}

		public double nextDouble() {
			if (st == null || !st.hasMoreTokens())
				try {
					readLine();
				} catch (Exception e) {
				}
			return Double.parseDouble(st.nextToken());
		}

		public Long nextLong() {
			if (st == null || !st.hasMoreTokens())
				try {
					readLine();
				} catch (Exception e) {
				}
			return Long.parseLong(st.nextToken());
		}

		public String next() {
			if (st == null || !st.hasMoreTokens())
				try {
					readLine();
				} catch (Exception e) {
				}
			return st.nextToken();
		}

		public static void readLine() {
			try {
				st = new StringTokenizer(br.readLine());
			} catch (Exception e) {
			}
		}
	}
}