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:
/*
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 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */

import java.io.*;
import java.util.*;
public class D1148 
{
    public static void main(String args[])throws IOException
    {
        Reader sc=new Reader();
        int n=sc.nextInt();
        int lim=2*n;
        List<Pair> less=new ArrayList<>();
        List<Pair> big=new ArrayList<>();
        for(int i=0;i<n;i++)
        {
            int x=sc.nextInt();
            int y=sc.nextInt();
            
            if(x<y)
                less.add(new Pair(x,y,i+1));
            else
                big.add(new Pair(x,y,i+1));
        }
        Collections.sort(big,new Comparator<Pair>(){
            public int compare(Pair ob1,Pair ob2)
            {
                return Integer.compare(ob1.b,ob2.b);
            }
        });
        Collections.sort(less,new Comparator<Pair>(){
            public int compare(Pair ob1,Pair ob2)
            {
                return Integer.compare(ob1.b,ob2.b)*-1;
            }
        });
        if(less.size()>big.size())
        {
            System.out.println(less.size());
            for(int i=0;i<less.size();i++)
            {
                System.out.print(less.get(i).ind+" ");
            }
        }
        else
        {
            System.out.println(big.size());
            for(int i=0;i<big.size();i++)
            {
                System.out.print(big.get(i).ind+" ");
            }
        }
        
    }
    static class Pair
    {
        int a,b,ind;
        public Pair(int aa,int bb,int i)
        {
            a=aa;
            b=bb;
            ind=i;
        }
    }
}

class Reader {
		final private int BUFFER_SIZE = 1 << 16;
		private DataInputStream din;
		private byte [] buffer;
		private int bufferPointer, bytesRead;
		
		public Reader () {
			din = new DataInputStream (System.in);
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}
		
		public Reader (String file_name) throws IOException {
			din = new DataInputStream (new FileInputStream (file_name));
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}
		
		public String readLine () throws IOException {
			byte [] buf = new byte[1024];
			int cnt = 0, c;
			while ((c = read ()) != -1) {
				if (c == '\n')
					break;
				buf[cnt++] = (byte) c;
			}
			return new String (buf, 0, cnt);
		}
		
		public int nextInt () throws IOException {
			int ret = 0;
			byte c = read ();
			while (c <= ' ')
				c = read ();
			boolean neg = (c == '-');
			if (neg)
				c = read ();
			do {
				ret = ret * 10 + c - '0';
			} while ((c = read ()) >= '0' && c <= '9');
			if (neg)
				return -ret;
			return ret;
		}
		
		public long nextLong () throws IOException {
			long ret = 0;
			byte c = read ();
			while (c <= ' ')
				c = read ();
			boolean neg = (c == '-');
			if (neg)
				c = read ();
			do {
				ret = ret * 10 + c - '0';
			} while ((c = read ()) >= '0' && c <= '9');
			if (neg)
				return -ret;
			return ret;
		}
		
		public double nextDouble () throws IOException {
			double ret = 0, div = 1;
			byte c = read ();
			while (c <= ' ')
				c = read ();
			boolean neg = (c == '-');
			if (neg)
				c = read ();
			do {
				ret = ret * 10 + c - '0';
			} while ((c = read ()) >= '0' && c <= '9');
			if (c == '.')
				while ((c = read ()) >= '0' && c <= '9')
					ret += (c - '0') / (div *= 10);
			if (neg)
				return -ret;
			return ret;
		}
		
		private void fillBuffer () throws IOException {
			bytesRead = din.read (buffer, bufferPointer = 0, BUFFER_SIZE);
			if (bytesRead == -1)
				buffer[0] = -1;
		}
		
		private byte read () throws IOException {
			if (bufferPointer == bytesRead)
				fillBuffer ();
			return buffer[bufferPointer++];
		}
		
		public void close () throws IOException {
			if (din == null)
				return;
			din.close ();
		}
	}