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>
#include<vector>
#include<cmath>

int main(){
	int n, x;
	std::cin>>n>>x;

	if(x==1 || x== 2*n-1){
		std::cout<<"No"<<std::endl;
		return 0;
	}

	std::cout<<"Yes"<<std::endl;

	int count1=0;
	int count2=0;
	if(x>=n)
		count2++;

	while(count1<n-1){
		if(n-count2 == x){
			count2++;
			continue;
		}

		std::cout<<n-count2<<std::endl;
		count1++;
		count2++;
	}

	std::cout<<x<<std::endl;
	count1=0;
	count2=0;

	while(count1<n-1){
		if((2*n-1)-count2 == x){
			count2++;
			continue;
		}

		std::cout<<(2*n-1)-count2<<std::endl;
		count1++;
		count2++;
	}

	return 0;
}