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 Rust (1.15.1) (judge verdict: Wrong Answer)

#![allow(unused_imports)]
#![allow(unused_macros)]
use std::cmp::{ min, max };

macro_rules! ewriteln {
    ($($args:expr),*) => { let _ = writeln!(&mut std::io::stderr(), $($args),*); };
}
macro_rules! trace {
    ($x:expr) => { ewriteln!(">>> {} = {:?}", stringify!($x), $x) };
    ($($xs:expr),*) => { trace!(($($xs),*)) }
}
macro_rules! put {
    ($x:expr) => { println!("{}", $x) };
    ($x:expr, $($xs:expr),*) => { print!("{} ", $x); put!($($xs),*) }
}

fn cut(a: usize, m: usize) -> usize {
    // cut m-th bit
    (a & ((1 << m) - 1)) | ((a >> (m + 1)) << m)
}

fn ins(a: usize, m: usize, b: usize) -> usize {
    // insert b as m-th bit
    ((a >> m) << (m + 1)) |
        (b << m) |
        (a & ((1 << m) - 1))
}

fn solve(n: usize, a: usize, b: usize) -> Option<Vec<usize>> {

    // trace!("solve", n, a, b);
    if (a ^ b).count_ones() % 2 == 0 { return None }

    if n == 1 {
        if a == b { return None }
        Some(vec![a, b])
    } else {
        for i in 0..n {
            let x = (a & (1 << i)) >> i;
            let y = (b & (1 << i)) >> i;
            if x == y { continue }
            let a2 = cut(a, i);
            let b2 = cut(b, i);
            for v in 0..(1 << (n-1)) {
                if v != a2 && v != b2 {
                    match (solve(n - 1, a2, v), solve(n - 1, v, b2)) {
                        (Some(xs), Some(ys)) => {
                            let mut res = vec![];
                            for &z in xs.iter() {
                                res.push(ins(z, i, x));
                            }
                            for &z in ys.iter() {
                                res.push(ins(z, i, y));
                            }
                            return Some(res)
                        },
                        _ => { return None }
                    }
                }
            }
        }
        None
    }
}

fn main() {
    let mut sc = Scanner::new();
    let n: usize = sc.cin();
    let a: usize = sc.cin();
    let b: usize = sc.cin();
    match solve(n, a, b) {
        Some(ans) => {
            put!("YES");
            for i in 0..ans.len() {
                if i < ans.len() - 1 {
                    print!("{} ", ans[i]);
                } else {
                    println!("{}", ans[i]);
                }
            }
        },
        None => {
            put!("NO");
        }
    }
}

use std::io::{self, Write};
use std::str::FromStr;
use std::collections::VecDeque;

struct Scanner { stdin: io::Stdin, buffer: VecDeque<String>, }
impl Scanner {
    fn new() -> Scanner { Scanner { stdin: io::stdin(), buffer: VecDeque::new() } }
    fn cin<T: FromStr>(&mut self) -> T {
        while self.buffer.len() == 0 {
            let mut line = String::new();
            let _ = self.stdin.read_line(&mut line);
            for w in line.split_whitespace() {
                self.buffer.push_back(String::from(w));
            }
        }
        self.buffer.pop_front().unwrap().parse::<T>().ok().unwrap()
    }
}