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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
long long int c1(long long int k)
{
	if (k <= 0)
		return 1;
	long long int res1[10], res2[10];
	res1[0] = res2[0] = 1;
	int i;
	for (i = 1; i < k; i++)
	{
		res1[i] = res2[i - 1];
		res2[i] = res2[i - 1] + res1[i - 1];
	}
	return res1[k - 1];
}
long long int c2(long long int k)
{
	if (k <= 0)
		return 1;
	long long int res1[10], res2[10];
	res1[0] = res2[0] = 1;
	int i;
	for (i = 1; i < k; i++)
	{
		res1[i] = res2[i - 1];
		res2[i] = res2[i - 1] + res1[i - 1];
	}
	return res2[k - 1];
}
int main()
{
	long long int h, w, k;
	scanf("%lld %lld %lld", &h, &w, &k);
	long long int p = 1000000007;
	long long int dp[102][10];
	k--;
	int i, j;
	for (i = 0; i <= h; i++)
		for (j = 0; j < w; j++)
			dp[i][j] = 0;
	dp[0][0] = 1;
	for (i = 0; i < h; i++)
	{
		for (j = 0; j < w; j++)
		{
			dp[i + 1][j] += dp[i][j] * c2(j) * c2(w - j - 1);
			dp[i + 1][j] %= p;
			if (j > 0)
				dp[i + 1][j] += dp[i][j - 1] * c1(j) * c2(w - j - 1);
			dp[i + 1][j] %= p;
			if (j < w - 1)
				dp[i + 1][j] += dp[i][j + 1] * c2(j) * c1(w - j - 1);
			dp[i + 1][j] %= p;
		}
	}
	printf("%lld\n", dp[h][k]);
	return 0;
}