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)

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}

const big_int: u64 = 1000_000_000 + 7;

fn main() {
    let v: Vec<usize> = read_vec();
    let h = v[0];
    let w = v[1];
    let g = v[2] - 1;
    let mut dp = vec![vec![0; w]; h + 1];
    dp[0][0] = 1;

    for i in 0..h {
        for j in 0..w {
            for k in 0..2u64.pow((w - 1) as u32) {
                let mut k = k;
                let mut line = vec![false; w - 1];
                let mut is_valid = true;
                for l in 0..w - 1 {
                    line[l] = if k % 2 == 1 {true} else {false};
                    k = k >> 1;
                    if l >= 1 {
                        if line[l] && line[l - 1] {
                            is_valid = false;
                            break;
                        }
                    }
                }
                if !is_valid {
                    continue;
                }
                if j >= 1 && line[j - 1] {
                    dp[i + 1][j - 1] += dp[i][j];
                    dp[i + 1][j - 1] %= big_int;
                } else if j < w - 1 && line[j] {
                    dp[i + 1][j + 1] += dp[i][j];
                    dp[i + 1][j + 1] %= big_int;
                } else {
                    dp[i + 1][j] += dp[i][j];
                    dp[i + 1][j] %= big_int;
                }
            }
        }
    }
    println!("{:?}", dp[h][g]);
}