Problem:
Score: 400 points

Problem Statement

Amidakuji is a traditional method of lottery in Japan.

To make an amidakuji, we first draw W parallel vertical lines, and then draw horizontal lines that connect them. The length of each vertical line is H+1 [cm], and the endpoints of the horizontal lines must be at 1, 2, 3, ..., or H [cm] from the top of a vertical line.

A valid amidakuji is an amidakuji that satisfies the following conditions:

No two horizontal lines share an endpoint.

The two endpoints of each horizontal lines must be at the same height.

A horizontal line must connect adjacent vertical lines.

Find the number of the valid amidakuji that satisfy the following condition, modulo 1\ 000\ 000\ 007: if we trace the path from the top of the leftmost vertical line to the bottom, always following horizontal lines when we encounter them, we reach the bottom of the K-th vertical line from the left.

For example, in the following amidakuji, we will reach the bottom of the fourth vertical line from the left.

Constraints

H is an integer between 1 and 100 (inclusive).

W is an integer between 1 and 8 (inclusive).

K is an integer between 1 and W (inclusive).

Input

Input is given from Standard Input in the following format:

H W K

Output

Print the number of the amidakuji that satisfy the condition, modulo 1\ 000\ 000\ 007.

Sample Input 1

1 3 2

Sample Output 1

1

Only the following one amidakuji satisfies the condition:

Sample Input 2

1 3 1

Sample Output 2

2

Only the following two amidakuji satisfy the condition:

Sample Input 3

2 3 3

Sample Output 3

1

Only the following one amidakuji satisfies the condition:

Sample Input 4

2 3 1

Sample Output 4

5

Only the following five amidakuji satisfy the condition:

Sample Input 5

7 1 1

Sample Output 5

1

As there is only one vertical line, we cannot draw any horizontal lines. Thus, there is only one amidakuji that satisfies the condition: the amidakuji with no horizontal lines.

Sample Input 6

15 8 5

Sample Output 6

437760187

Be sure to print the answer modulo 1\ 000\ 000\ 007.

Solution in Rust (1.15.1) (judge verdict: Accepted)

#![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),*) }
}

const M: u64 = 1_000_000_007;

fn main() {
    let mut sc = Scanner::new();
    let h: usize = sc.cin();
    let w: usize = sc.cin();
    let k: usize = sc.cin();

    let mut memo = vec![vec![0_u64; w]; h + 1];
    memo[0][0] = 1;

    let mut table = vec![-1; w];

    for i in 1..h+1 {
        for p in 0..1<<(w-1) {
            let edges: Vec<usize> = (0..w-1).filter(|j| p & (1 << j) > 0).collect();
            let mut valid = true;
            if edges.len() >= 2 {
                for i in 0..edges.len() - 1 {
                    if edges[i] + 1 == edges[i+1] {
                        valid = false;
                        break;
                    }
                }
            }
            if !valid { continue }
            for u in 0..w { table[u] = 0; }
            for &u in edges.iter() {
                table[u] = 1;
                table[u+1] = -1;
            }
            for u in 0..w {
                memo[i][u] +=
                    match table[u] {
                        1 => memo[i-1][u+1],
                        -1 => memo[i-1][u-1],
                        _ => memo[i-1][u]
                    };
                memo[i][u] %= M;
            }
        }
    }
    put!(memo[h][k-1]);
}

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()
    }
}