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! 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)*}
    };
}
 
#[allow(unused_macros)]
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)*}
    };
}
 
#[allow(unused_macros)]
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>>()
    };
 
    ($next:expr, bytes) => {
        read_value!($next, String).into_bytes()
    };
 
    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };
 
    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}









use std::cmp::min;
use std::collections::BTreeMap;
use std::process;
fn main() {

    input! {
        H: usize,
        W: usize,
        K: usize
    }
    let modu = 1000000007;
    if W == 1{
        println!("1");
        process::exit(0);
    }
    let mut dp = vec![vec![0; W+1]; H+1 ];
    dp[0][0] = 1;

    //let valid_horizontal: Vec<_> = (0..(1<<(W-1))).filter(|a| {(0..W).all(|i|{ 3 != ((a>>i)&3)})}).collect();
    for j  in 1..((H+1) as i64){
        for k  in 0..(W as i64){
            let mut L:u64 = 0;
            let mut C:u64 = 0;
            let mut R:u64 = 0;
            for b in 0..1<<(W-1){
                let mut flag = 0;
                for bb in 0..(W-2){
                    //println!("{:#02b}", b);
                    if ((b>>bb)&3 == 3){
                        flag = 1;
                        break;    
                    }
                }
                if flag == 1{continue;}
                if k-1 >= 0 && (W-2) as i64 >= k{
                    if (b>>(k-1))&1 == 1{
                        L += 1;
                    }
                    else if (b>>k)&1 == 1{
                        R += 1;
                    }
                    else{
                        C += 1;
                    }
                }
                else if k-1<0{
                    if (b>>k)&1 == 1{
                        R+= 1;
                    }
                    else{
                        C+= 1;
                    }
                }
                else{
                    if (b>>(k-1))&1 == 1{
                        L += 1;
                    }
                    else{
                        C += 1;
                    }
                }
            }
            let jj= j as usize;
            let kk = k as usize;
            dp[jj][kk] = (dp[jj][kk] + (dp[jj-1][kk]*C))% modu;
            if k-1>= 0{
                dp[jj][kk] = (dp[jj][kk] + (dp[jj-1][kk-1] * L))%modu;
            }
            if ((W-2) as i64)>= k{
                dp[jj][kk] = (dp[jj][kk] + (dp[jj-1][kk+1] * R))%modu;
            }

        }
    }
    println!("{}",dp[H][K-1]);   
}