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.List;
import java.util.Scanner;
import java.util.stream.Collectors;
import java.util.stream.IntStream;

public class Main {

    public static void main(String[] args) {
        Scanner stdin = new Scanner(System.in);
        int n = stdin.nextInt();
        int x = stdin.nextInt();

        if (x == 1 || x == 2 * n - 1) {
            System.out.println("No");
            System.exit(0);
        }

        List<Integer> digits = IntStream.range(2, 2 * n - 1).filter(v -> v != x).boxed().collect(Collectors.toList());


        int len = digits.size();
        int mid = len / 2;

        System.out.println("Yes");
        for (int i = 0; i < mid; i++) System.out.println(digits.get(i));
        System.out.println(1);
        System.out.println(x);
        System.out.println(2 * n - 1);
        for (int i = mid; i < len; i++) System.out.println(digits.get(i));
    }

}