Problem:
Score : 400 points

Problem Statement

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.

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.

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.

Sample Input 1

4 4

Sample Output 1

Yes
1
6
3
7
4
5
2

This case corresponds to the figure in the problem statement.

Sample Input 2

2 1

Sample Output 2

No

No matter what permutation was written into the blocks of step N, the integer written into the block of step 1 would be 2.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
#define tr(container, it) \
        for (auto it = container.begin(); it != container.end(); it++)
#define scontains(c, x) ((c).find(x) != (c).end())   //O(log n)
#define contains(c, x) (find((c).begin(),(c).end(),x) != (c).end()) //O(n)
#define ill(_x)  ll _x;scanf("%lld",&_x);
#define idb(_x)  double _x;scanf("%lf",&_x);
#define all(x) (x).begin(),(x).end()
#define pll pair<ll,ll>
#define mll map<ll,ll>
#define vll vector<ll>
#define sll set<ll>
#define vs vector<string>
#define in0(x, a, b)((x)>=a && (x)<=b    )
#define in1(x, a, b)((x)>a && (x)<b)
#define  rep(i, begin, end) for (__typeof(end) i = (begin) - ((begin) > (end)); i != (end) - ((begin) > (end)); i += 1 - 2 * ((begin) > (end)))
const double pi = 3.14159265358979323846;
const int INF = 0x3f3f3f3f;
const int X10 = 1024, X11 = 2048, X12 = 4096, X13 = 8196, X14 = 16392, X15 = 32786, X16 = 65536, X17 = 131072, X18 = 262144, X19 = 524288, X20 = 1048576;
const int MOD = (int)(1e9+7);

class UnionFind {
private:
    vector<int> p, rank;
    int N;
    int findSet2(int i, bool* A){A[i] = 1;return (p[i] == i) ? i : (p[i] = findSet2(p[i], A));}
public:
    UnionFind(int N) {
        rank.assign(N, 0);
        p.assign(N, 0);
        for (int i = 0; i < N; i++) p[i] = i;
        this->N = N;
    }

    int findSet(int i) { return (p[i] == i) ? i : (p[i] = findSet(p[i])); }

    bool isSameSet(int i, int j) { return findSet(i) == findSet(j); }

    void unionSet(int i, int j) {
        if (!isSameSet(i, j)) {
// if from different set
            int x = findSet(i), y = findSet(j);
            if (rank[x] > rank[y]) p[y] = x;
// rank keeps the tree short
            else {
                p[x] = y;
                if (rank[x] == rank[y]) rank[y]++;
            }
        }
    }


    int numDisjointSets(){
        set<int> s;
        bool* visited = new bool[this->N];
        memset(visited, 0, sizeof(bool)*this->N);
        for (int i = 0; i < N; ++i) {
            if(!visited[i])s.insert(findSet2(i, visited));
        }
        return s.size();
    }

    int maxElements(){
        int mx = 0;
        map<int, int> m;
        for (int i = 0; i < N; ++i) {
            m[findSet(i)] ++;
        }
        set<pair<int, int>> s;
        for(auto p : m){
            s.insert({p.second, p.first});
        }
        return s.rbegin()->first ;
    }


};

void _(){
    int n,x;
    cin >> n >> x;
    if(n==x){
        cout << "Yes\n";
        for (int i = 0; i < 2*n-1; ++i) {
            cout << i+1 << "\n";
        }
    }
    else if(abs(x-n)==1 && n>2){
        if(n==3){
            if(x==2)cout << "1\n3\n2\n4\n5";
            else cout << "3\n5\n4\n1\n2";
        }
        else {
            for (int i = 0; i < x - 3; ++i) {
                cout << i+1 <<"\n";
            }
            cout << x-2 << "\n" << x+2 <<"\n" << x << "\n" << x-1 <<"\n" << x+1 << "\n";
            for (int i = x+2; i < 2*n-1; ++i) {
                cout << i+1 <<"\n";
            }
        }
    }
    else cout <<"No\n";




}


int main() {
#ifdef Debug
    freopen("/home/joduskame/Desktop/cpp/IO/Input.txt", "r", stdin);
    freopen("/home/joduskame/Desktop/cpp/IO/Output.txt", "w", stdout);
#endif
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    _();
    return 0;

}