Problem: We have a pyramid with N steps, built with blocks. The steps are numbered 1 through N from top to bottom. For each 1≤i≤N, step i consists of 2i-1 blocks aligned horizontally. The pyramid is built so that the blocks at the centers of the steps are aligned vertically.

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A pyramid with N=4 steps

Snuke wrote a permutation of (1, 2, ..., 2N-1) into the blocks of step N. Then, he wrote integers into all remaining blocks, under the following rule:

* The integer written into a block b must be equal to the median of the three integers written into the three blocks directly under b, or to the lower left or lower right of b.



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Writing integers into the blocks

Afterwards, he erased all integers written into the blocks. Now, he only remembers that the integer written into the block of step 1 was x.

Construct a permutation of (1, 2, ..., 2N-1) that could have been written into the blocks of step N, or declare that Snuke's memory is incorrect and such a permutation does not exist.

Constraints

* 2≤N≤10^5
* 1≤x≤2N-1

Input

The input is given from Standard Input in the following format:


N x


Output

If no permutation of (1, 2, ..., 2N-1) could have been written into the blocks of step N, print `No`.

Otherwise, print `Yes` in the first line, then print 2N-1 lines in addition.

The i-th of these 2N-1 lines should contain the i-th element of a possible permutation.

Examples

Input

4 4


Output

Yes
1
6
3
7
4
5
2


Input

2 1


Output

No

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.*;

public class Main{

	public static void main(String[] args) throws IOException {
		MyScanner sc = new MyScanner(System.in);
		int n=sc.nextInt();
		int x = sc.nextInt();
		if(x==1||x==2*n-1) {
			System.out.println("No");
			return;
		}
		System.out.println("Yes");
		if(n==2) {
			System.out.println(1);
			System.out.println(2);
			System.out.println(3);
			return;
		}
		if(x+1!=2*n-1) {
			Queue<Integer> q=new ArrayDeque<>();
			for(int i=2; i<2*n-1;i++) {
				if(i!=x+1&&i!=x)
					q.add(i);
			}
			for(int i=0; i<n-3; i++) {
				int e=q.poll();
				System.out.println(e);
			}
			System.out.println(x+1);
			System.out.println(1);
			System.out.println(x);
			System.out.println(2*n-1);
			for(int i=0; i<n-2; i++) {
				System.out.println(q.poll());
			}	
		} else {
			Queue<Integer> q=new ArrayDeque<>();
			for(int i=2; i<2*n-1;i++) {
				if(i!=x-1&&i!=x)
					q.add(i);
			}
			for(int i=0; i<n-3; i++) {
				int e=q.poll();
				System.out.println(e);
			}
			System.out.println(x-1);
			System.out.println(1);
			System.out.println(x);
			System.out.println(2*n-1);
			for(int i=0; i<n-2; i++) {
				System.out.println(q.poll());
			}	
		}
	}

	static class MyScanner
	{
		BufferedReader br;
		StringTokenizer st;
		public MyScanner(InputStream s) 
		{
			br=new BufferedReader(new InputStreamReader(s));
		}
		public String nextLine() throws IOException
		{
			return br.readLine();
		}
		public String next() throws IOException
		{
			while(st==null || !st.hasMoreTokens())
				st=new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		public int nextInt() throws IOException
		{
			return Integer.parseInt(next());
			
		}
		public double nextDouble() throws IOException
		{
			return Double.parseDouble(next());
		}
		public boolean ready() throws IOException
		{
			return br.ready();
		}
		public long nextLong() throws IOException
		{
			return Long.parseLong(next());
		}
	}
}