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<stdio.h>
#include<algorithm>
using namespace std;
int n, x, w[201000], Loc[201000];
void Put(int a, int b){
    swap(w[a], w[Loc[b]]);
    swap(Loc[w[a]], Loc[w[Loc[b]]]);
}
int main(){
    int i;
    scanf("%d%d",&n,&x);
    if(x==1||x==2*n-1){
        printf("No\n");
        return 0;
    }
    for(i=1;i<=n*2-1;i++)w[i]=i, Loc[i]=i;
    if(x==2){
        Put(n,x);
        Put(n-1,1);
    }
    else{
        Put(n,x);
        Put(n-1,x-1);
        Put(n+1,x+1);
        if(n>=2)Put(n+2,x-2);
    }
    printf("Yes\n");
    for(i=1;i<=n*2-1;i++)printf("%d ",w[i]);
    printf("\n");
}