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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <iomanip>
#include <cmath>
#include <vector>
#include <queue>
#include <limits>
#include <algorithm>
#include <string>
#include <cstring>
#include <map>

using namespace std;
typedef long long ll;
const int MOD = 1000000007;
#define MAX_N (100010)

int main()
{
	int H, W, K;
	scanf("%d%d%d", &H, &W, &K);

	ll dp[110][10];

	memset(dp, 0, sizeof(dp));
	dp[0][1] = 1;
	for (int h = 1; h <= H; h++) {
		for (int w = 1; w <= W; w++) {
			for (int i = 0; i < (1 << (W - 1)); i++) {
				bool prevIsOn = false;
				bool valid = true;
				for (int j = 0; j < (W - 1); j++) {
					if (i & (1 << j)) {
						if (prevIsOn) {
							valid = false;
							break;
						}
						prevIsOn = true;
					} else {
						prevIsOn = false;
					}
				}
				if (!valid)
					continue;

				if (w - 2 >= 0 && (i & (1 << (w - 2)))) {
					dp[h][w] += dp[h - 1][w - 1];
				}
				if (w - 1 <= W - 2 && i & (1 << (w - 1))) {
					dp[h][w] += dp[h - 1][w + 1];
				}
				if ((w == 1 && (i & (1 << 0)) == 0) ||
					(w == W && (i & (1 << (w - 2))) == 0) ||
					((w - 2 >= 0 && (i & (1 << (w - 2))) == 0) && (w - 1 <= W - 2 && (i & (1 << (w - 1))) == 0))) {
					dp[h][w] += dp[h - 1][w];
				}
				while (dp[h][w] >= MOD)
					dp[h][w] -= MOD;
			}
		}
	}

	printf("%lld\n", dp[H][K]);

	return 0;
}