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)

// Try AtCoder
// author: Leonardone @ NEETSDKASU

fn main() {
	let mut stdin = String::new();
    std::io::Read::read_to_string(
    	&mut std::io::stdin(),
        &mut stdin).unwrap();
	let mut stdin = stdin.split_whitespace();
    let mut get = || stdin.next().unwrap();
    macro_rules! get {
    	($t:ty) => (get().parse::<$t>().unwrap());
        () => (get!(i64));
    }
    
    let n = get!();
    let a = get!();
    let b = get!();
    
    let mut xs = vec![];
    let mut ai = 0;
    let mut bi = 0;
    for i in 0..1<<n {
    	let x = i ^ (i >> 1);
        xs.push(x);
        if x == a {
        	ai = i;
        }
        if x == b {
        	bi = i;
        }
    }
    
    if ai - bi == 1 {
    	println!("YES");
        for i in 0..xs.len() {
        	if i > 0 {
            	print!(" ");
            }
        	let i = (ai as usize + i) % xs.len();
            print!("{}", xs[i]);
        }
        println!();
        return;
    }
    if bi - ai == 1 {
    	println!("YES");
        for i in 0..xs.len() {
        	if i > 0 {
            	print!(" ");
            }
        	let i = (ai as usize + xs.len() - i) % xs.len();
            print!("{}", xs[i]);
        }
        println!();
        return;
    }
	
    print!("NO");
}