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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#define fo(i,a,b) for(int i=(a);i<(b);i++)
#define MOD 1000000007
#define MP make_pair
#define PB push_back
typedef long long ll;

int N, x, a[200200];
vector<int> v;

int main () {
        scanf("%d %d", &N, &x);
        if (x == 1 || x == 2*N-1) return puts("No"), 0;
        puts("Yes");
        a[N-2] = 1;
        a[N-1] = x;
        a[N] = 2*N-1;
        fo(i, 2, 2*N-1) if (i != x) v.PB(i);
        fo(i, 0, 2*N-1) if (i<N-2 || i>N) a[i] = v.back(), v.pop_back();
        fo(i, 0, 2*N-1) printf("%d\n", a[i]);
        return 0;
}