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: Wrong Answer)

import java.util.*;

public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int h = sc.nextInt();
    int w = sc.nextInt();
    int k = sc.nextInt() - 1;
    long MOD = (long)Math.pow(10, 9) + 7;
    long[][][] dp = new long[w][w][h + 1];
    if(w == 1) {
      System.out.println(1);
    } else {
      long r1 = (long)Math.pow(2, w - 2);
      long r2 = 1;
      if(w >= 3) r2 = (long)Math.pow(2, w - 3);
      dp[0][0][1] = r1;
      dp[0][1][1] = r2;
      dp[w - 1][w - 1][1] = r1;
      dp[w - 1][w - 2][1] = r2;
      for(int i = 1; i < w - 1; i++) {
        dp[i][i][1] = r2;
        dp[i][i - 1][1] = r2;
        dp[i][i + 1][1] = r2;
      }
      for(int s = 2; s <= h; s++) {
        for(int i = 0; i < w; i++) {
          for(int j = 0; j < w; j++) {
            if(i == 0) {
              dp[0][j][s] = (r1 * dp[0][j][s - 1]) % MOD;
              long t = (r2 * dp[1][j][s - 1]) % MOD;
              dp[0][j][s] = (dp[0][j][s] + t) % MOD;
            } else if(i == (w - 1)) {
              dp[w - 1][j][s] = (r1 * dp[w - 1][j][s - 1]) % MOD;
              long t = (r2 * dp[w - 2][j][s - 1]) % MOD;
              dp[w - 1][j][s] = (dp[w - 1][j][s] + t) % MOD;              
            } else {
              long t1 = (r2 * dp[i][j][s - 1]) % MOD;
              long t2 = (r2 * dp[i - 1][j][s - 1]) % MOD;
              long t3 = (r2 * dp[i + 1][j][s - 1]) % MOD;
              dp[i][j][s] = (t1 + t2) % MOD;
              dp[i][j][s] = (dp[i][j][s] + t3) % MOD;
            }
          }
        }
      }
      System.out.println(dp[0][k][h]);
    }
  }
}