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.

<image>

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.



<image>

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
#include <bits/stdc++.h>
using namespace std;

int main(){
  int n,x,len,mid;

  cin>>n>>x;
  len=n*2-1;
  mid=len/2;
  int ans[200001]={},cnt;
  if(x==1||len==x){cout<<"No"<<endl;return 0;}

  if(x==2){
    ans[mid]=x;
    ans[mid-1]=1;
    cnt=3;
  }
  else if(x==len-1){
    ans[mid]=x;
    ans[mid+1]=len;
    cnt=1;
  }
  else{
    ans[mid-1]=x;
    ans[mid-2]=1;
    ans[mid]=len;
    ans[mid+1]=2;
    cnt=3;
  }

  for(int i=0;i<len;i++){
    if(cnt==x)cnt++;
    if(!ans[i])ans[i]=cnt++;
  }
  cout<<"Yes"<<endl;
  for(int i=0;i<len;i++)cout<<ans[i]<<endl;
      
  return 0;
}
