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:
from collections import deque

N,x = list(map(int,input().split()))

if x==2*N-1 or x==1:
    print("No")
else:
    Q=deque()
    if x<=N:
        Q.append(x)
        Q.append(x-1)
        for i in range(2*N-1,0,-1):
            if i!=x and i!=x-1:
                if i%2==0:
                    Q.append(i)
                else:
                    Q.appendleft(i)
    else:
        Q.append(x)
        Q.append(x+1)
        for i in range(1,2*N):
            if i!=x and i!=x+1:
                if i%2==0:
                    Q.append(i)
                else:
                    Q.appendleft(i)

    print("Yes")

    while len(Q)>0:
        print(Q.pop())