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 <iostream>
#include <vector>
#include <math.h>
#include <cmath>
#include <algorithm>
#include <numeric>
#include <string>
#include <cstring>
#include <regex>
#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using pii = pair<int,int>;
using pl4 = pair<ll,ll>;
using vi = vector<int>;
using vvi = vector<vi>;
using vs = vector<string>;
using vvs = vector<vs>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vpii = vector<pii>;
using vvpii = vector<vpii>;
using vpl4 = vector<pl4>;
using vvpl4 = vector<vpl4>;

#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define pob pop_back()
#define sz size()
#define be begin()
#define en end()
#define asn assign
#define emp empty()
#define ft front()
#define bk back()
#define clr clear()
#define ins insert
#define ers erase
#define res resize

#define FOR(i,a,b) for(int i=(a);i<=(b);i++)
#define rFOR(i,a,b) for(int i=(b);i>=(a);i--)
#define REP(i,a) for(int i=0;i<(a);i++)
#define REP1(i,a) for(int i=1;i<=(a);i++)
#define rREP(i,a) for(int i=(a-1);i>=0;i--)
#define rREP1(i,a) for(int i=(a);i>0;i--)
#define SORT(a) sort((a).be,(a).en)
#define rSORT(a) sort((a).rbegin(),(a).rend())
#define UNIQUE(a) (a).erase(unique((a).be,(a).en),(a).en)
#define PREVP(a) prev_permutation((a).be,(a).en)
#define NEXTP(a) next_permutation((a).be,(a).en)
#define BINS(a,b) binary_search((a).be,(a).en,(b))
#define LOWB(a,b) (lower_bound((a).be,(a).en,(b))-(a).be)
#define UPB(a,b) (upper_bound((a).be,(a).en,(b))-(a).be)
#define CNT(a,b) count((a).be,(a).en,b)
#define SUM(a) accumulate((a).be,(a).en,0)
#define REV(a) reverse((a).be,(a).en)
#define REGS(a,b) regex_search((a),regex(b))
#define REGM(a,b) regex_match((a),regex(b))
#define yn(a) cout <<((a)?"yes":"no")<<endl;
#define Yn(a) cout <<((a)?"Yes":"No")<<endl;
#define YN(a) cout <<((a)?"YES":"NO")<<endl;
#define say(a) cout <<(a);
#define sal(a) cout <<(a)<<endl;
#define sak cout <<endl;
#define dbg(a) cout <<(#a)<<": "<<(a)<<endl;
#define a2l(a) ((ll)(a-97))
#define A2l(a) ((ll)(a-65))
#define l2a(a) ((char)(a+97))
#define l2A(a) ((char)(a+65))
#define DigN2(a) ((llabs(a)==0)?(1):((ll)(log2(double(llabs(a))))+1))
#define DigN10(a) ((llabs(a)==0)?(1):((ll)(log10(double(llabs(a))))+1))
#define Dig2(a,b) (((a)>>(b))&1)
#define Dig10(a,b) (ll)(((a)/((ll)(pow(10.0,(double)(b)))))%10)
#define Pow2(a) (1<<(a))
#define llin(a) ll (a);cin >>(a);
#define stin(a) string (a);cin >>(a);
#define rdn(a,b) ((a)/(b))
#define rou(a,b) ((((double(a)/double(b))-((a)/(b)))<0.5)?((a)/(b)):(((a)/(b))+1))
#define rup(a,b) ((((a)%(b))==0)?((a)/(b)):(((a)/(b))+1))
#define min(a,b) ((a<b)?(a):(b))
#define max(a,b) ((a>b)?(a):(b))

#define int ll

const ll MOD = 1e9+7;
const string alp = "abcdefghijklmnopqrstuvwxyz";
const string ALP = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

ll gcd(ll a, ll b){if(b==0)return a;return gcd(b,a%b);}
ll lcm(ll a, ll b){return a/gcd(a,b)*b;}

ll sgn(ll n){
    if(n==0) return 0;
    else return n/abs(n);
}

ll salv(vll v){
  say("{");
  FOR(i,0,v.sz-1){
    say(v[i]);
    if(i!=v.sz-1) say(",");
  }
  sal("}")
}

ll DigS10(ll n){
  ll m=0;
  FOR(i,0,DigN10(n)-1){
    m+=(ll)((llabs(n)%(ll)(pow(10.0,(double)(i+1))))/(ll)(pow(10.0,(double)i)));
  }
  return m;
}

ll isP(ll n){
  if(n<=1) return 0;
  FOR(i,2,(ll)sqrt(n)){
    if(n%i==0) return 0;
  }
  return 1;
}

vll FactM(1,1);
vll FactMI(1,1);

ll PowM(ll a,ll b){
  ll ans=1,x=(a%MOD);
  FOR(i,0,DigN2(b)-1){
    if(Dig2(b,i)==1) ans=(ans*x)%MOD;
    if(i!=(DigN2(b)-1)) x=(x*x)%MOD;
  }
  return ans;
}

void CFactM(ll n){
  FOR(i,FactM.sz,n){
    FactM.pb((FactM[i-1]*(i%MOD))%MOD);
  }
  return;
}

void CFactMI(ll n){
  CFactM(n);
  if(FactMI.sz<(n+1)) FactMI.res(n+1,-1);
  if(FactMI[n]==-1) FactMI[n]=PowM(FactM[n],MOD-2);
  rFOR(i,1,n-1){
    if(FactMI[i]!=-1) break;
    FactMI[i]=((FactMI[i+1]*((i+1)%MOD))%MOD);
  }
  return;
}

ll CombM(ll n,ll k){
    if((n<0)||(k<0)) return 0;
    if(n<k) return 0;
    if(n+1>FactMI.sz) CFactMI(n);
    return ((((FactMI[k]*FactMI[n-k])%MOD)*FactM[n])%MOD);
}

signed main() {
  ll n,x,m;
  cin >> n >> x;
  m=n*2-1;
  if(x!=m&&x!=1){
    cout << "Yes" << endl;
    if(n==2){
      cout << 1 << endl << 2 << endl << 3;
      return 0;
    }
    if (x==2) {
      REP(i,n-2){
        cout << 5+i << endl;
      }
      cout << 3 << endl << 1 << endl << 2 << endl << 4 << endl;
      REP(i,n-3){
        cout << m-i << endl;
      }
    } else {
      ll a=0;
      REP(i,n-2){
        if(3+i==x)a++;
        cout << 3+i+a << endl;
      }
      cout << 2 << endl << m << endl << x << endl << 1 << endl;
      a=0;
      REP(i,n-3){
        if(m-i==x)a++;
        cout << m-i-a << endl;
      }
    }
  } else {
    cout << "No";
  }
  
}