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:
N,x = map(int,input().split())
if x == 1:
  print('No')
elif x == 2*N-1:
  print('No')
else:
  print('Yes')
  if N == 2:
    for i in range(1,4):
      print(i)
  else:
    if x != 2*N-2:
      L = []
      M = [x+2,x-1,x,x+1]
      for i in range(1,2*N):
        if not i in M:
          L.append(i)
      for j in range(N-2):
        print(L[j])
      for k in range(4):
        print(M[k])
      for l in range(N-3):
        print(L[l+N-2])
    else:
      L = []
      M = [x-2,x,x+1,x-1]
      for i in range(1,2*N):
        if not i in M:
          L.append(i)
      for j in range(N-2):
        print(L[j])
      for k in range(4):
        print(M[k])
      for l in range(N-3):
        print(L[l+N-2])