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 <iostream>
using namespace std;
int main(void){
    int n, x;
    cin >> n >> x;
    
    if (x == 1 || x == 2 * n - 1) cout << "No" << endl;
    else {
        cout << "Yes" << endl;
        if (n == x) {
            for (int i = 0; i < 2 * n - 1; i++) cout << i + 1 << endl;
        }
        else if (n < x) {
            for (int i = 0; i < 2 * n - 1; i++)
            cout << (i + x - n) % (2*n - 1) + 1 << endl;
        }
        else {
            for (int i = 0; i < 2 * n - 1; i++)
            cout << (i + (2 * n - 1) - n + x) % (2*n - 1) + 1 << endl;
        }
    }
    
    
}