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)

use std::io::prelude::*;

fn input<T: std::str::FromStr>() -> T {
    let stdin = std::io::stdin();
    let token: String = stdin
        .lock()
        .bytes()
        .map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().unwrap()
}

#[allow(dead_code)]
fn with_stdout_lock<F: FnOnce(std::io::BufWriter<std::io::StdoutLock>)>(f: F) {
    let out = std::io::stdout();
    let w = std::io::BufWriter::new(out.lock());
    f(w);
}

#[allow(unused_macros)]
macro_rules! debug {
    ($var:expr) => {
        match &$var { v => eprintln!(concat!(stringify!($var), "={:?}"), v) }
    };
    ($var:expr, $($vars:expr),*) => {
        match &$var { v => eprint!(concat!(stringify!($var), "={:?} "), v) }
        debug!($($vars),*);
    };
}

fn main() {
    let h: usize = input();
    let w: usize = input();
    let k: usize = input();
    const MOD: i64 = 1000000007;

    if w == 1 {
        println!("1");
        return;
    }

    let b: Vec<usize> = (0..w).collect();
    let mut dp = vec![vec![0i64; w]; h + 1];
    dp[0][0] = 1;
    for i in 0..h {
        for j in 0..1 << (w - 1) {
            let ok = b
                .windows(2)
                .all(|b| (j >> b[0]) & 1 == 0 || (j >> b[1]) & 1 == 0);
            if !ok {
                continue;
            }

            let mut to: Vec<usize> = (0..w).collect();
            for k in 0..w - 1 {
                if (j >> k) & 1 == 1 {
                    to.swap(k, k + 1);
                }
            }

            for k in 0..w {
                dp[i + 1][to[k]] += dp[i][k];
                dp[i + 1][to[k]] %= MOD;
            }
        }
    }

    println!("{}", dp[h][k - 1]);
}