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>
#include<algorithm>
using namespace std;
int n,x,vis[200005],a[200005],hd=1,i;
void pt(int x,int y)
{
	vis[x]=1;
	a[y]=x;
}
int main()
{
	scanf("%d%d",&n,&x);
	if(x==1||x==2*n-1)
	{
		printf("No\n");
		return 0;
	}
	if(n==2)
	{
		printf("Yes\n1\n2\n3\n");
		return 0;
	}
	printf("Yes\n");
	vis[x]=1;vis[1]=vis[2*n-1]=1;
	a[n]=x;a[n+1]=2*n-1;a[n-1]=1;
	if(x==2*n-2)pt(x-1,n+2);
	else pt(x+1,n-2);
	hd=1;
	for(i=1;i<=2*n-1;i++)
	{
		if(a[i])printf("%d\n",a[i]);
		else
		{
			while(vis[hd])hd++;
			printf("%d\n",hd);
			vis[hd]=1;
		}
	}
	return 0;
}