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)

#pragma region
#define _USE_MATH_DEFINES
#include <iostream>
#include <string>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <map>
#include <queue>
#include <stack>
#include <set>
#include <list>
#include <iomanip>
#include <cstdint>
#include <bitset>

using namespace std;
typedef long long ll;
//#define rep(i, s, e) for (int(i) = (s); (i) < (e); ++(i))
#define rep(i, e) for (int(i) = 0; (i) < (e); ++(i))
#define rrep(i, s) for (int(i) = (s) - 1; (i) >= 0; --(i))
#define all(x) x.begin(),x.end()

#pragma region UnionFind
struct UnionFind
{
	vector<int> par;

	UnionFind(int n) : par(n, -1) {}
	void init(int n) { par.assign(n, -1); }

	int root(int x)
	{
		if (par[x] < 0) return x;
		else return par[x] = root(par[x]);
	}

	bool issame(int x, int y)
	{
		return root(x) == root(y);
	}

	bool merge(int x, int y)
	{
		x = root(x); y = root(y);
		if (x == y) return false;
		if (par[x] > par[y]) swap(x, y);
		par[x] += par[y];
		par[y] = x;
		return true;
	}

	int size(int x)
	{
		return -par[root(x)];
	}
};
#pragma endregion
#pragma region GCD
int gcd(int a, int b)
{
	if (a%b == 0)return b;
	return gcd(b, a%b);
}
#pragma endregion
#pragma region chmin
template<class T> inline bool chmin(T& a, T b)
{
	if (a > b)
	{
		a = b;
		return true;
	}
	return false;
}
#pragma endregion
#pragma region chmax
template<class T> inline bool chmax(T& a, T b)
{
	if (a < b)
	{
		a = b;
		return true;
	}
	return false;
}
#pragma endregion
#pragma endregion


int main()
{
	int h, w, k; cin >> h >> w >> k; --k;
	vector<int> v;
	rep(i, pow(2, w - 1))
	{
		bool ok = true;
		rep(j, w - 2)if (i&(1 << j) && i&(1 << j + 1))ok = false;
		if (ok)v.push_back(i);
	}
	vector<vector<int>> dp(h + 1, vector<int>(w));
	dp[0][0] = 1;
	int MOD = 1e9 + 7;
	rep(i, h)rep(j, w)
	{
		for (int x : v)
		{
			if (j > 0 && x&(1 << (j - 1)))
			{
				dp[i + 1][j - 1] += dp[i][j];
				dp[i + 1][j - 1] %= MOD;
			}
			else if (j < w - 1 && x&(1 << j))
			{
				dp[i + 1][j + 1] += dp[i][j];
				dp[i + 1][j + 1] %= MOD;
			}
			else
			{
				dp[i + 1][j] += dp[i][j];
				dp[i + 1][j] %= MOD;
			}
		}
	}
	cout << dp[h][k] << endl;
}