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>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;

int main(){
	int n,x; scanf("%d%d",&n,&x);
	if(x==1 || x==2*n-1) puts("No");
	else{
		puts("Yes");
		vector<int> sign(2*n-1);
		if(x<=n){
			fill(sign.begin(),sign.end(),1);
			rep(i,x-1) sign[n+i]=-1;
		}
		else{
			fill(sign.begin(),sign.end(),-1);
			rep(i,(2*n-1)-x) sign[n+i]=1;
		}
		sign[n-1]=0;

		int large=x+1,small=x-1;
		rep(i,2*n-1){
			if     (sign[i]>0) printf("%d\n",large), large++;
			else if(sign[i]<0) printf("%d\n",small), small--;
			else               printf("%d\n",x);
		}
	}

	return 0;
}