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 <cstdio>

using namespace std;

const int N = 2e5 + 500;

int n, x, p[N], c = 1;

int main(){
	scanf("%d%d", &n, &x);
	if(x == 1 || x == 2 * n - 1){
		printf("No\n");
		return 0;
	}
	if(x <= n){
		p[n - 1] = x;
		p[n] =     x - 1;
		p[n - 2] = x + 1;
		p[n + 1] = x + 2;
	}
	else{
		p[n - 1] = x;
		p[n] =     x + 1;
		p[n - 2] = x - 1;
		p[n + 1] = x - 2;
	}
	printf("Yes\n");
	for(int i = 0;i < 2 * n - 1;i++){
		if(!p[i]){
			if((x <= n && c == x - 1))
				c = x + 3;
			if((x > n && c == x - 2))
				c = x + 2;
			p[i] = c++;
		}
		printf("%d\n", p[i]);
	}
}