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)

using namespace std;
#include <iostream>
#include <bits/stdc++.h>

#define infile "../test/sample-1.in"
#define int long long
// #define INF 2147483647
#define INF 1000000000000000000LL
#define MOD 1000000007LL

// {{{ define basic macro
#define _overload3(_1,_2,_3,name,...) name
#define _rep(i,n) repi(i,0,n)
#define repi(i,a,b) for(int i=(int)(a);i<(int)(b);++i)
#define rep(...) _overload3(__VA_ARGS__,repi,_rep,)(__VA_ARGS__)
#define _rrep(i,n) repi(i,0,n)
#define rrepi(i,a,b) for(int i=(int)(b-1);i>=(int)(a);--i)
#define rrep(...) _overload3(__VA_ARGS__,rrepi,_rrep,)(__VA_ARGS__)

#define foreach(x,a) for(auto& (x) : (a) )
#define each(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++)

#define all(x) (x).begin(),(x).end()
#define sum(v) accumulate(all(v), 0)
#define sz(x) ((int)(x).size())
template<class T> inline void chmax(T &a, const T &b) { if(a < b) a = b; }
template<class T> inline void chmin(T &a, const T &b) { if(a > b) a = b; }
// n次元配列の初期化。第２引数の型のサイズごとに初期化していく。
template<typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val){
    std::fill( (T*)array, (T*)(array+N), val );
}
#define fill(x,y) memset(x,y,sizeof(x))
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define uni(x) sort(all(x));x.erase(unique(all(x)),x.end())
#define ten(n) ((int)1e##n)

template <class T = int>
T in() {T x; cin>>x; return (x);}

struct Fast {
  Fast(){
    std::cin.tie(0);
    ios::sync_with_stdio(false);
  }
} fast;
// }}}
//{{{ dump macro
#ifdef PCM
  #include "dump.hpp"
#else
  #define dump(...) 42
  #define dump_1d(...) 42
  #define dump_2d(...) 42
#endif
//}}}
//{{{ others
typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef long double ld;
typedef pair<int,int> pii;
typedef tuple<int,int,int> iii;
template<typename T>
using PQ = priority_queue<T, vector<T>, greater<T>>;

int dx[]={1, -1, 0, 0};
int dy[]={0, 0, 1, -1};
//}}}


int solve(){
  int n,x;cin>>n>>x;

  if (x!=n) {
    cout << "No" << endl;
  }
  else{
    rep(i, 1, 2*n){
      cout << i << endl;
    }
  }

  return 0;
}

int test(){ //{{{
  return 0;
}
//}}}

signed main() { //{{{
#ifdef INPUT_FROM_FILE
  std::ifstream in(infile);
  std::cin.rdbuf(in.rdbuf());
#endif
#ifdef PCM
  test();
#endif
  solve();
  return 0;
} //}}}