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;

int P[int(2e5+5)]={};
bool used[int(2e5+5)];

int main() {
	int N,x;
	scanf("%d%d",&N,&x);
	if(x == 1 || x == 2*N-1) {
		printf("No\n");
		return 0;
	}
	printf("Yes\n");
	P[N] = x, used[x] = true;
	P[N+1] = x+1, used[x+1] = true;
	P[N-1] = x-1, used[x-1] = true;
	if(x > 2) P[N+2] = x-2, used[x-2] = true;
	if(x < 2*N-2) P[N-2] = x+2, used[x+2] = true;
	int r = 0;
	for(int i = 1; i <= 2*N-1; ++i) {
		if(P[i] != 0) continue;
		++r;
		while(used[r]) ++r;
		P[i] = r;
	}
	for(int i = 1; i <= 2*N-1; ++i)
		printf("%d\n",P[i]);
}