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.Scanner;

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

    int N = scanner.nextInt();
    int x = scanner.nextInt();
    if (x == 1 || x == N * 2 - 1) System.out.println("No");
    else {
      System.out.println("Yes");
      int[] a = new int[N * 2 - 1];
      int c = 2;
      for (int i = 0; i < N - 2; i++) {
        if (c == x) c++;
        a[i] = c++;
      }
      a[N - 2] = 1;
      a[N - 1] = x;
      a[N] = N * 2 - 1;
      for (int i = N + 1; i < N * 2 - 1; i++) {
        if (c == x) c++;
        a[i] = c++;
      }

      StringBuilder sb = new StringBuilder();
      for (int i : a) sb.append(i).append('\n');
      System.out.print(sb.toString());
    }
  }

  private static int process(int[] a) {
    while (a.length > 1) {
      int[] b = new int[a.length - 2];
      for (int i = 1; i <= b.length; i++) {
        b[i - 1] = median(a[i - 1], a[i], a[i + 1]);
      }
      a = b;
    }
    return a[0];
  }

  private static int median(int a, int b, int c) {
    if (a > b) return median(b, a, c);
    if (b > c) return median(a, c, b);
    return b;
  }
}