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 PYTHON3):

Solution:
import sys

read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines

N, n = map(int, read().split())

def main(N, n):
    if n in (1, 2 * N - 1):
        print('No')
        return
    if N == 2:
        A = [1, 2, 3]
        print('Yes')
        print(*A, sep='\n')
        return
    if n == 2:
        a, b, c, d = 3, 2, 1, 4
    else:
        a, b, c, d = n - 1, n, n + 1, n - 2
    nums = list(range(1, N + N))
    for x in (a, b, c, d):
        nums.remove(x)
    k = len(nums) // 2
    A = nums[:k] + [a, b, c, d] + nums[k:]
    print('Yes')
    print(*A, sep='\n')

main(N, n)