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_macros)]
macro_rules! get {
    ([$t:ty] ; $n:expr) => {
        (0..$n)
        .map(|_| get!([$t])).collect::<Vec<_>>()
    };
    ($($t:ty),+ ; $n:expr) => {
        (0..$n).map(|_| get!($($t),+)).collect::<Vec<_>>()
    };
    ([$t:ty]) => {
        rl().split_whitespace()
        .map(|x| x.parse().unwrap())
        .collect::<Vec<$t>>()
    };
    ($t:ty) => {
        rl().parse::<$t>().unwrap()
    };
    ($($t:ty),*) => {
        {
            let line =rl();
            let mut iter = line.split_whitespace();
            ($(iter.next().unwrap().parse::<$t>().unwrap()),*)
        }
    };
}
fn rl() -> String {
    let mut line = String::new();
    std::io::stdin().read_line(&mut line).unwrap();
    line.trim().to_string()
}

const MOD: usize = 1000000007;

fn f(bit: usize, w: u32) -> bool {
    for i in 0..w - 2 {
        let a = 2_usize.pow(w - 2 - i);
        let b = 2_usize.pow(w - 2 - i - 1);
        if bit & a == a && bit & b == b {
            return false;
        }
    }
    true
}

fn main() {
    let (h, w, k) = get!(usize, usize, usize);
    if w == 1 {
        println!("1");
        return;
    }
    let mut dp = vec![vec![0; w + 1]; h + 1];
    dp[0][0] = 1;
    for i in 0..h {
        for j in 0..w {
            for bit in 0..1 << (w - 1) {
                if !f(bit, w as u32) {
                    continue;
                }
                //eprintln!("bit: {:08b}", bit);
                if 1 <= j && bit >> j - 1 & 1 == 1 {
                    dp[i + 1][j - 1] += dp[i][j];
                    dp[i + 1][j - 1] %= MOD;
                } else if j <= w - 2 && bit >> j & 1 == 1 {
                    dp[i + 1][j + 1] += dp[i][j];
                    dp[i + 1][j + 1] %= MOD;
                } else {
                    dp[i + 1][j] += dp[i][j];
                    dp[i + 1][j] %= MOD;
                }
            }
        }
    }
    //for i in 0..h {
    //    eprintln!("{:?}", dp[i]);
    //}
    //eprintln!("{:?}", dp);
    println!("{}", dp[h][k - 1]);
}