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>
#define REP(i,n) for(int i=0;i<(int)(n);i++)
#define ALL(x) (x).begin(),(x).end()
#define LL long long

using namespace std;

int main(){

    int N;cin>>N;
    int x;cin>>x;

    int ans[2*N-1];
    if(x==1||x==2*N-1)cout<<"No"<<endl;
    else{
        ans[N-1]=x;
        ans[N-2]=x-1;
        ans[N]  =x+1;

        int cnt =2*N-1;
        REP(i,2*N-1){
            if(i>=N-2&&i<=N)continue;

            while(cnt>=x-1&&cnt<=x+1)cnt--;
            ans[i]=cnt--;
        }
        cout<<"Yes"<<endl;
        REP(i,2*N-1)cout<<ans[i]<<endl;

    }


    return 0;
}