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>
using namespace std;

const int maxn=1e5+10;
int n,x,ans[maxn<<1];
stack<int> s,l;

int main(){
	cin>>n>>x;
	if(x==1||x==2*n-1){
		puts("No");
		return 0;
	}
	puts("Yes");
	for(int i=1;i<x;++i)
		s.push(i);
	for(int i=x+1;i<2*n;++i)
		l.push(i);
	ans[n]=x;
	for(int i=n-1;i;--i)
		if(s.empty()){
			ans[i]=l.top();
			l.pop();
		}
		else{
			ans[i]=s.top();
			s.pop();
		}
	for(int i=n+1;i<2*n;++i)
		if(l.empty()){
			ans[i]=s.top();
			s.pop();
		}
		else{
			ans[i]=l.top();
			l.pop();
		}
	for(int i=1;i<2*n;++i)
		printf("%d\n",ans[i]);
	return 0;
}