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: Wrong Answer)

use std::str::FromStr;
use std::fmt::Debug;

fn read_as_vec<T>() -> Vec<T>
where T: FromStr, <T as FromStr>::Err : Debug{
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().split_whitespace().map(|c| T::from_str(c).unwrap()).collect()

}

fn read_as_string() -> String{
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().to_string()
}


// between $target and $target+1
fn enumerate(w: u64, target: u64, previous: bool) ->u64{
    if w ==1 {
        return 1;
    }

    
    if !previous {
        if target == w-1{
            return enumerate(w-1, target, true);
        }else{
            return enumerate(w-1, target, true) + enumerate(w-1, target, false);
        }
    }else{
        if target == w-1{
            return 0;
        }else{
            return enumerate(w-1, target, false);
        }
    }
}

fn main(){
    let line : Vec<usize> = read_as_vec();
    let (h, w, k) = (line[0], line[1], line[2]);
    let mut kuji: Vec<Vec<u64>> = vec![vec![0;w];h+1];
    kuji[0][0] = 1;

    let e = enumerate(5, 2, false);

    if w == 1{
        println!("1");
        return;
    }
    for i in 1..h+1{
        for j in 0..w{
//            println!("i = {}, j = {}", i, j);
            if j == 0{
                let straight = enumerate(w as u64, 0, false)
                - enumerate(w as u64, (j+1) as u64, false);
                kuji[i][j] = kuji[i-1][j+1] * enumerate(w as u64, (j+1) as u64, false) +
                    kuji[i-1][j] * straight;
            }else if j == w -1{
                let straight = enumerate(w as u64, 0, false)
                    - enumerate(w as u64, (j+1) as u64, false);
                kuji[i][j] = kuji[i-1][j-1] * enumerate(w as u64, j as u64, false) +
                    kuji[i-1][j] * straight;
            }else{
                let straight = enumerate(w as u64, 0, false)
                - enumerate(w as u64, j as u64, false)
                - enumerate(w as u64, (j+1) as u64, false);
                //go to left
                kuji[i][j] = kuji[i-1][j-1] * enumerate(w as u64, j as u64, false) +
                    kuji[i-1][j+1] * enumerate(w as u64, (j+1) as u64, false) +
                    kuji[i-1][j] * straight;
            }
            kuji[i][j] %= 1000000007;
        }
    }

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