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 math
import fractions
import heapq
import collections
import re
import array
import bisect

from collections import Counter, defaultdict


II = lambda: int(input())
MI = lambda: map(int, input().split())


def main():
    N, x = MI()
    if x in [1, 2 * N - 1]:
        print("No")
        return
    print("Yes")
    if N == 2:
        print("1\n2\n3")
        return
    s = set([i for i in range(1, 2 * N)])
    a = [None] * (2 * N - 1)
    MED = N - 1
    if x == 2 * N - 2:
        l = (x - 1, x, x + 1, x - 2)
        a[MED - 1:MED + 3] = l
        for i in l:
            s.remove(i)
    else:
        l = x + 1, x, x - 1, x + 2
        a[MED - 1:MED + 3] = l
        for i in l:
            s.remove(i)
    for i in range(2 * N - 1):
        if a[i] is None:
            a[i] = s.pop()
    for i in range(2 * N - 1):
        print(a[i])


if __name__ == "__main__":
    main()