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 PYTHON):

Solution:
N,x = map(int, raw_input().split())

if x==1 or x==2*N-1:
    print "No"
else:
    print "Yes"
    if N==2:
        T = [1,2,3]
    else:
        T = range(1, 2*N)
        T[N-1] = x
        if x<N:
            T[N-3] = 2*N-1
            T[N-2] = 1
            T[N]   = 2*N-2
        else:
            T[N-3] = 1
            T[N-2] = 2*N-1
            T[N]   = 2
        S = set(range(1, 2*N))
        for t in T[N-3:N+1]:
            S.remove(t)
        for i in range(N-3):
            T[i] = S.pop()
        for i in range(N+1, 2*N-1):
            T[i] = S.pop()
    for t in T:
        print t