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 <bits/stdc++.h>
#define int long long
#define mod (int)(1e9+7)
#define inf (int)(3e18+7)
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,n) for(int i=1;i<n;i++)
#define P pair<int,int>
#define all(v) v.begin(),v.end()
#define mkp make_pair
#define mkt make_tuple
#define prique(T) priority_queue<T,vector<T>,greater<T>>
#define vecunique(vec) sort(vec.begin(), vec.end());decltype(vec)::iterator result = std::unique(vec.begin(), vec.end());vec.erase(result, vec.end())
using namespace std;

bool prime(int x) {
	for (int i = 2; i * i <= x; i++) {
		if (x % i == 0)return false;
	}
	return x > 1;
}
int gcd(int x, int y) {
	if (y == 0)return x;
	return gcd(y, x % y);
}
int lcm(int x, int y) {
	return x * y / gcd(x, y);
}
int kai(int x) {
	if (x == 0)return 1;
	return kai(x - 1) * x % mod;
}
int mod_pow(int x, int y, int mod_) {
	int res = 1;
	while (y > 0) {
		if (y & 1) {
			res = res * x % mod_;
		}
		x = x * x % mod_;
		y >>= 1;
	}
	return res;
}
int comb(int x, int y) {
	return kai(x)* mod_pow(kai(x - y), mod - 2, mod) % mod * mod_pow(kai(y), mod - 2, mod) % mod;
}
/*--------Library Zone!--------*/

int h, w, k;
int dp[114][19];
signed main() {
	cin >> h >> w >> k;
	dp[0][0] = 1;
	rep(i, h) {
		rep(j, w) {
			dp[i][j] %= mod;
			rep(k, 1 << (w - 1)) {
				rep(l, w - 2) {
					if ((k >> l) & 1 && (k >> (l + 1) & 1))goto ng;
				}
				if (j && (k >> (j - 1)) & 1)dp[i + 1][j - 1] += dp[i][j];
				else if ((k >> j) & 1)dp[i + 1][j + 1] += dp[i][j];
				else dp[i + 1][j] += dp[i][j];
			ng:;
			}
		}
	}
	cout << dp[h][k - 1] % mod << endl;
}