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.

<image>

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.



<image>

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
#include <iostream>
using namespace std;
int p[200005],n,x;
bool used[200005];
void set(int i,int j)
{
	if (j<1 || j>=2*n || used[j])
	return;
	p[i]=j;
	used[j]=1;
}
int main()
{
	scanf("%d%d",&n,&x);
	if (x==1 || x==2*n-1)
	{
		printf("No");
		return 0;
	}
	printf("Yes\n");
	if (n==2)
	{
		printf("1 2 3");
		return 0;
	}
	set(n,x);
	set(n-1,x-1);
	set(n+1,x+1);
	set(n-2,x+2);
	set(n+2,x-2);
	int cur=1;
	for (int i=1;i<2*n;i++)
	{
		if (!p[i])
		{
			while (used[cur])
			cur++;
			set(i,cur);
		}
		printf("%d ",p[i]);
	}
}