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 main()
{
	int N, x;
	cin >> N >> x;

	string ans = "No";
	int n = 2 * N - 1;
	vector<int> v( n );
	if( 1 < x && x < n ) {
		ans = "Yes";
		int n2 = n / 2;
		v[n2 - 1] = x - 1;
		v[n2] = x;
		v[n2 + 1] = x + 1;
		int j = 1;
		for( int i = 0; i < n2 - 1; i++ ) {
			if( j == x - 1 ) j = x + 2;
			v[i] = j;
			j++;
		}
		for( int i = n2 + 2; i < n; i++ ) {
			if( j == x - 1 ) j = x + 2;
			v[i] = j;
			j++;
		}
	}

	cout << ans << endl;
	if( ans == "Yes" ) {
		for( int i = 0; i < n; i++ ) {
			cout << v[i] << endl;
		}
	}
}