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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)


import java.util.Scanner;

public class Main {

    static final long MOD = 1000000007;
    
    static int H, W, K;
    static long[][] dp = new long[110][10];
    
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        H = sc.nextInt();
        W = sc.nextInt();
        K = sc.nextInt() - 1;
        dp[0][0] = 1;
        for(int h=1; h<=H; ++h) {
            for(int w=0; w<W; ++w) {
                int[] kuji = new int[W-1];
                //mannaka
                if(w < W-1) kuji[w] = -1;
                if(w-1 >= 0) kuji[w-1] = -1;
                dp[h][w] += (kumi(kuji, 0) * dp[h-1][w]) % MOD;
                dp[h][w] %= MOD;
                //left
                if(w > 0) {
                    kuji = new int[W-1];
                    kuji[w-1] = 2;
                    dp[h][w] += (kumi(kuji, 0) * dp[h-1][w-1]) % MOD;
                    dp[h][w] %= MOD;
                }
                if(w < W-1) {
                    kuji = new int[W-1];
                    kuji[w] = 2;
                    dp[h][w] += (kumi(kuji, 0) * dp[h-1][w+1]) % MOD;
                    dp[h][w] %= MOD;
                }
            }
        }
        System.out.println(dp[H][K]);
    }
    
    static long kumi(int[] kuji, int n) {
        if(n == W-1) {
            for(int i=0; i<kuji.length-1; ++i) {
                if(kuji[i] > 0 && kuji[i+1] > 0) {
                    return 0;
                }
            }
            return 1;
        }
        long pos = 0;
        if(kuji[n] > 1) pos = kumi(kuji, n+1);
        else if(kuji[n] == 0) {
            kuji[n] = 1;
            pos = kumi(kuji, n+1);
            kuji[n] = 0;
        }
        long neg = 0;
        if(kuji[n] < 2) neg = kumi(kuji, n+1);
        return (neg + pos) % MOD;
    }
}