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;

signed main(){
	int a,b;cin>>a>>b;
	if(b==1||b==2*a-1){puts("No");return 0;}
	puts("Yes");
	if(a==2){puts("1");puts("2");puts("3");}
	else{
		vector<int>ans(2*a-1);
		set<int>L;
		for(int i=1;i<2*a;i++)L.insert(i);
		ans[a-1]=b;
		ans[a-2]=b-1;
		ans[a]=b+1;
		L.erase(b);
		L.erase(b-1);
		L.erase(b+1);
		if(b==2){
			ans[a-3]=b+2;
			L.erase(b+2);
			}
		else{
			ans[a+1]=b-2;
			L.erase(b-2);
			}	
		for(int i=0;i<2*a-1;i++){
			if(ans[i]==0){
				ans[i]=*L.begin();
				L.erase(L.begin());
				}
			}	
		for(int i:ans){
			cout<<i<<endl;
			}
		}
	}