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)

// Try AtCoder
// author: Leonardone @ NEETSDKASU

fn main() {
	let mut stdin = String::new();
    std::io::Read::read_to_string(
    	&mut std::io::stdin(),
        &mut stdin).unwrap();
	let mut stdin = stdin.split_whitespace();
    let mut get = || stdin.next().unwrap();
    macro_rules! get {
    	($t:ty) => (get().parse::<$t>().unwrap());
        () => (get!(i64));
    }
    
    let h = get!(usize);
    let w = get!(usize);
    let k = get!(usize);
    
    // use std::collections::HashMap;
    let md = 1_000_000_007;
    let mut dp = vec![0_i64; w];
    
    let mut tb = vec![];
    for i in 0..(1<<(w-1)) {
        let mut ok = true;
        for j in 0..w {
            let v = 3 << j;
            if i & v == v {
                ok = false;
                break;
            }
        }
        if ok {
            tb.push(i);
            // for j in 0..w { print!("{}", (i>>j)&1); }
            // println!();
            
        }
    }
    
    dp[0] = 1;
    for _ in 0..h {
    	let mut tmp = vec![0; w];
        for i in 0..w {
        	if 0 < i && i+1 < w {
                let mut c = 0;
                let b = 3 << (i-1);
                for &t in tb.iter() {
                    if t&b==0 { c += 1; }
                }
            	let e = dp[i]*c%md;
        		tmp[i] = (tmp[i]+e) % md;
            } else {
                let mut c = 0;
                let b = if i==0 { 1 } else { 1<<(i-1) };
                for &t in tb.iter() {
                    if t&b==0 { c += 1; }
                }
            	let e = dp[i]*c%md;
        		tmp[i] = (tmp[i]+e) % md;
            }
            if i+1 < w {
            	if i==0 || i+2==w {
                    let mut c = 0;
                    let b = if i==0 { 3 } else { 3 << (i-1) };
                    let v = if i==0 { 1 } else { 2 << (i-1) };
                    for &t in tb.iter() {
                        if t & b == v { c += 1; }
                    }
	            	let e = dp[i] * c % md;
    	    		tmp[i+1] = (tmp[i+1]+e) % md;
                } else {
                    let mut c = 0;
                    let b = 7 << (i-1);
                    let v = 2 << (i-1);
                    for &t in tb.iter() {
                        if t&b==v { c += 1; }
                    }
                  	let e = dp[i]*c%md;
                  	tmp[i+1] = (tmp[i+1]+e) % md;
                }
            }
            if i > 0 {
            	if i==1 || i+1==w {
                    let mut c = 0;
                    let b = if i==1 { 3 } else { 3 << (i-1) };
                    let v = if i==1 { 1 } else { 1 << (i-1) };
                    for &t in tb.iter() {
                        if t&b==v { c += 1; }
                    }
	            	let e = dp[i]*c%md;
    	    		tmp[i-1] = (tmp[i-1]+e) % md;
                } else {
                    let mut c = 0;
                    let b = 7 << (i-2);
                    let v = 2 << (i-2);
                    for &t in tb.iter() {
                        if t&b==v { c += 1; }
                    }
                  	let e = dp[i]*c%md;
                  	tmp[i-1] = (tmp[i-1]+e) % md;
                }
            }
        }
        dp = tmp;
        // println!("{:?}", dp);
    }
    println!("{}", dp[k-1]);
    
}