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 (1.42.0) (judge verdict: Wrong Answer)

use std::io;
use std::str::FromStr;
use std::cmp::max;

static MOD: u64 = 1_000_000_007;

fn input_tuple() -> (usize, usize, usize) {
    let mut line: String = String::new();
    io::stdin().read_line(&mut line);
    let vec: Vec<usize> = line.split_whitespace().map(|s| s.parse().unwrap()).collect();
    (vec[0], vec[1], vec[2])
}

fn binary_power(x: u64, n: u64) -> u64 {
    if n == 0 {
      return 1;
    }
    if n % 2 == 0 {
      return (binary_power(x, n / 2) % MOD).pow(2) % MOD;
    } else {
      return binary_power(x, n - 1) % MOD * x % MOD;
    }
}

fn main() {
    let (h, w, k) = input_tuple();
    let mut dp = vec![vec![0; w]; h+1];
    dp[0][0] = 1;
    for i in 1..(h+1) {
        for j in 0..w {
            if j > 0 {
                let mut left = max(w as i64 - 2, 0);
                if j < w - 1 {
                    left = max(left - 1, 0);
                }
                if j > 1 {
                    left = max(left - 1, 0);
                }
                dp[i][j] += dp[i-1][j-1] * binary_power(2, left as u64);
                dp[i][j] %= MOD;
            }

            let mut left = max(w as i64 - 1, 0);
            if j > 0 {
                left = max(left - 1, 0);
            }
            if j < w - 1 {
                left = max(left - 1, 0);
            }
            dp[i][j] += dp[i-1][j] * binary_power(2, left as u64);
            dp[i][j] %= MOD;

            if j + 1 < w {
                let mut left = max(w as i64 - 2, 0);
                if j + 1 < w - 1 {
                    left = max(left - 1, 0);
                }
                if j + 1 > 1 {
                    left = max(left - 1, 0);
                }
                dp[i][j] += dp[i-1][j+1] * binary_power(2, left as u64);
                dp[i][j] %= MOD;
            }
        }
    }
    // println!("{:?}", dp);
    println!("{}", dp[h][k - 1]);
}