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:
# coding:utf-8

import sys

INF = float('inf')
MOD = 10 ** 9 + 7

def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x) - 1 for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def II(): return int(sys.stdin.readline())
def SI(): return input()


def main():
    n, x = LI()

    if x == 1 or x == 2 * n - 1:
        print('No')
        return 0

    print('Yes')
    if n < 3:
        print(*[n for n in range(1, 2 * n)], sep='\n')
        return 1

    # n - 1段目の中心2つがxになるように並べる
    if x == 2:
        tmp = [3, 2, 1, 4]
    else:
        tmp = [x - 1, x, x + 1, x - 2]

    remain = list(set([n for n in range(1, 2 * n)]) - set(tmp))
    k = 3 - n
    tmp = remain[:k] + tmp + remain[k:]
    print(*tmp, sep='\n')

    return 1


main()