Problem:
Score : 800 points

Problem Statement

You are given integers N,\ A and B.
Determine if there exists a permutation (P_0,\ P_1,\ ...\ P_{2^N-1}) of (0,\ 1,\ ...\ 2^N-1) that satisfies all of the following conditions, and create one such permutation if it exists.

P_0=A

P_{2^N-1}=B

For all 0 \leq i < 2^N-1, the binary representations of P_i and P_{i+1} differ by exactly one bit.

Constraints

1 \leq N \leq 17

0 \leq A \leq 2^N-1

0 \leq B \leq 2^N-1

A \neq B

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N A B

Output

If there is no permutation that satisfies the conditions, print NO.

If there is such a permutation, print YES in the first line.
Then, print (P_0,\ P_1,\ ...\ P_{2^N-1}) in the second line, with spaces in between.
If there are multiple solutions, any of them is accepted.

Sample Input 1

2 1 3

Sample Output 1

YES
1 0 2 3

The binary representation of P=(1,0,2,3) is (01,00,10,11), where any two adjacent elements differ by exactly one bit.

Sample Input 2

3 2 1

Sample Output 2

NO

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

//{{{
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

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

using PII = pair<int, int>;
using TI3 = tuple<int, int, int>;
using LL = long long;
using ULL = unsigned long long;
using MAT = array<array<LL, 2>, 2>;

template<class T> constexpr inline T begin(const T&) {return 0;}
template<class T> constexpr inline T end(const T& x) {return max(T(0), x);}

#define SZ(x) int((x).size())
#define PB push_back
#define EB emplace_back
#define GET_MACRO(_1,_2,_3,_4,NAME,...) NAME
#define CHECK(a,b) static_assert(std::is_same<decltype(a), decltype(b)>::value, "REP diff types");

#define REPPP(i,s,c,t) for(i; ((s) and (c)) or (putchar("\n "[c]) and (c)); (t))
#define REPP2(i,n) REPPP(auto i=begin(n),i==begin(n),i!=end(n),++i)
#define REPP3(i,m,n) CHECK(m,n) REPPP(auto i=(m),i==(m),i<(n),++i)
#define REPP4(i,m,n,s) CHECK(m,n) REPPP(auto i=(m),i==(m),((s)>0 and i<(n)) or ((s)<0 and i>(n)),i+=(s))
#define REPP(...) GET_MACRO(__VA_ARGS__, REPP4, REPP3, REPP2)(__VA_ARGS__)

#define REP2(i,n) for(auto i=begin(n);i!=end(n);++i)
#define REP3(i,m,n) CHECK(m,n) for(auto i=(m);i<(n);++i)
#define REP4(i,m,n,s) CHECK(m,n) for(auto i=(m);((s)>0 and i<(n)) or ((s)<0 and i>(n));i+=(s))
#define REP(...) GET_MACRO(__VA_ARGS__, REP4, REP3, REP2)(__VA_ARGS__)

template<class T> void _read(T &x) {cin >> x;}
void _read(size_t &x) {scanf("%zu", &x);}
void _read(int &x) {scanf("%d", &x);}
void _read(LL &x) {scanf("%lld", &x);}
void _read(ULL &x) {scanf("%llu", &x);}
void _read(double &x) {scanf("%lf", &x);}
void _read(char &x) {scanf(" %c", &x);}
void _read(char *x) {scanf("%s", x);}
void read() {}
template<class T, class... U>
void read(T& head, U&... tail) {
    _read(head);
    read(tail...);
}

template<class T> void _print(const T &x) {cout << x;}
void _print(const size_t &x) {printf("%zu", x);}
void _print(const int &x) {printf("%d", x);}
void _print(const LL &x) {printf("%lld", x);}
void _print(const ULL &x) {printf("%llu", x);}
void _print(const double &x) {printf("%.16lf", x);}
void _print(const char &x) {putchar(x);}
void _print(const char *x) {printf("%s", x);}
void _print(const string &x) {printf("%s", x.c_str());}
template<class T> void _print(const vector<T> &x) {
    for(auto i = x.begin(); i != x.end(); _print(*i++)) {
        if (i != x.cbegin()) putchar(' ');
    }
}
void print() {putchar('\n');}
template<class T, class... U> void print(const T& head, const U&... tail) {
    _print(head);
    if(sizeof...(tail)) putchar(' ');
    print(tail...);
}


#define LLINF 0x3f3f3f3f3f3f3f3f
#define INF 0x3f3f3f3f

const int MOD = 1000000007;

//}}}
#define MAX_N 300005

void pprint(int x, int N) {
    REPP(i, N - 1, -1, -1) {
        _print((x >> i) & 1);
    }
}

void solve(int N, int A, int B, vector<int>& ret) {
    if(N == 1) {
        ret.push_back(A);
        ret.push_back(B);
        return;
    }

    int C = A ^ B;
    REP(i, N) {
        if(((C >> i) & 1) == 1) {
            int AA = ((A >> (i + 1)) << i) | (A & ((1 << i) - 1));
            int BB = ((B >> (i + 1)) << i) | (B & ((1 << i) - 1));
            vector<int> ret2;
            solve(N - 1, AA, AA ^ 1, ret);
            solve(N - 1, AA ^ 1, BB, ret2);

            REP(j, ret.size()) {
                int p = (A >> i) & 1;
                int x = ret[j], y = ret2[j];
                ret[j] = ((x >> i) << (i + 1)) | (p << i) | (x & ((1 << i) - 1));
                ret2[j] = ((y >> i) << (i + 1)) | ((p ^ 1) << i) | (y & ((1 << i) - 1));
            }
            ret.insert(ret.end(), ret2.begin(), ret2.end());
            break;
        }
    }
}

int main()
{
    int N, A, B;
    read(N, A, B);
    if((__builtin_popcount(A ^ B) & 1) == 0) {
        print("NO");
        return 0;
    }
    vector<int> ans;
    solve(N, A, B, ans);
    print("YES");
    print(ans);
    return 0;
}