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.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {

     static Scanner scanner;

     public static void main(String[] args) {
         scanner = new Scanner(System.in);

         int N=gi();
         int x=gi();
         int[] a=new int[2*N-1];
         List<Integer> f=new ArrayList<Integer>();

         if (x==1 || x==2*N-1) {
        	 System.out.println("No");
        	 return;
         } else {
        	 for(int i=0; i<2*N-1; i++) {
        		 a[i]=-1;
        	 }
        	 a[N-1]=x;
        	 a[N-2]=1;
        	 a[N]=2*N-1;
        	 f.add(x);
        	 f.add(1);
        	 f.add(2*N-1);
        	 if(N!=3 && x!=2) {
        		 a[N+1]=2;
        		 f.add(2);
        	 }
         }

         int ind=0;
         int v=1;
         while(ind<2*N-1) {
        	 if(a[ind]==-1) {
        		 while(f.contains(v)) {
        			 v++;
        		 }
        		 a[ind]=v;
        		 v++;
        		 ind++;
        	 } else {
        		 ind++;
        	 }
         }

         System.out.println("Yes");
         for(int i=0; i<2*N-1; i++) {
        	 System.out.println(a[i]);
    	 }
         //System.out.println(ans);
     }

     // 文字列として入力を取得
     public static String gs() {
          return scanner.next();
     }

     // intとして入力を取得
     public static int gi() {
          return Integer.parseInt(scanner.next());
     }

     // longとして入力を取得
     public static long gl() {
          return Long.parseLong(scanner.next());
     }

     // doubleとして入力を取得
     public static double gd() {
          return Double.parseDouble(scanner.next());
     }
}