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;
#define inf 1e9
#define ll long long 
#define M 1000000007
#define P pair<int,int>
#define FOR(i,m,n) for(int i=m;i<n;i++)
#define rep(i,n) FOR(i,0,n)
const int vx[4] = {0,1,0,-1};
const int vy[4] = {1,0,-1,0};
#define PI 3.14159265

int n;
int x;

int main(){

  cin>>n>>x;
  if(x==1||x==2*n-1){
    cout<<"No"<<endl;
  }
  else{
    cout<<"Yes"<<endl;
    vector<int> ans(2*n-1);
    rep(i,2*n-1){
      ans[(n+i-1)%(2*n-1)]=(x+i-1)%(2*n-1)+1;
    }
    rep(i,2*n-1){
      cout<<ans[i]<<endl;
    }
  }


  return 0;
}