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)

macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

fn main() {
    input!{
        h: usize,    // 縦棒の長さ - 1
        w: usize,    // 縦棒の本数
        k: usize,    // 最終的に辿り着かなければならない縦棒は左から何番目か
    }

    let m = 1_000_000_007;

    // a[h][x] h = 上からh列目を通過した直後, x = 左からx番目の縦棒にいる
    // その場合の数
    let mut a = [[0u64; 1000]; 1000];

    a[0][0] = 1;

    for h in 1..(h+1) {
        for bars in 0u8..(1 << (w-1)) {
            // 連結があるならスキップ
            if bars & (bars << 1) != 0 {
                continue;
            }

            for x in 0..w {
                let left = x > 0 && (bars & (1 << (x-1)) != 0);
                let right = x < w-1 && (bars & (1 << x) != 0);

                // println!("h = {}, x = {}", h, x);
                // println!("bar = {:b}, left = {}, right = {}", bars, left, right);
                // 左へ行くか
                if left {
                    a[h][x-1] += a[h-1][x];
                    a[h][x-1] %= m;
                }
                // 右へ行くか
                else if right {
                    a[h][x+1] += a[h-1][x];
                    a[h][x+1] %= m;
                }
                // まっすぐ降りるか
                else {
                    a[h][x] += a[h-1][x];
                    a[h][x] %= m;
                }

                // for x in 0..w {
                    // print!("{} ", a[h][x]);
                // }
                // println!("");

            }
        }

        // println!("h = {}, result ...", h);
        // for x in 0..w {
            // print!("{} ", a[h][x]);
        // }
        // println!("");
    }

    println!("{}", a[h][k-1] % m);
}