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 rep(i,n) for(int i = 0; i < (n); ++i)
using namespace std;
typedef long long int ll;
typedef pair<int, int> P;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

const int mod = 1000000007;
struct mint{
	ll x;
	mint(ll x = 0):x((x%mod+mod)%mod){}
	mint operator-() const{return mint(-x);}
	mint& operator+=(const mint a){
		if((x += a.x) >= mod) x -= mod;
		return *this;
	}
	mint& operator-=(const mint a){
		if((x += mod-a.x) >= mod) x -= mod;
		return *this;
	}
	mint& operator*=(const mint a){
		(x *= a.x) %= mod;
		return *this;
	}
	mint operator+(const mint a) const{
		mint res(*this);
		return res+=a;
	}
	mint operator-(const mint a) const{
		mint res(*this);
		return res-=a;
	}
	mint operator*(const mint a) const{
		mint res(*this);
		return res*=a;
	}
	mint pow(ll t) const{
		if(!t) return 1;
		mint a = pow(t>>1);
		a *= a;
		if(t&1) a *= *this;
		return a;
	}
	mint inv() const{
		return pow(mod-2);
	}
	mint& operator/=(const mint a){
		return (*this) *= a.inv();
	}
	mint operator/(const mint a) const{
		mint res(*this);
		return res/=a;
	}
	bool operator==(const mint a){
		return x == a.x;
	}
};

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	int H, W, K;
	cin >> H >> W >> K;
	vector<vector<mint>> dp(H+1, vector<mint>(W, 0));
	dp[0][0] = 1;
	rep(i,H){
		rep(j,W){
			rep(k,(1<<(W-1))){
				bool ok = true;
				rep(l,W-2){
					if(((k>>l) & 1) && ((k>>(l+1)) & 1)) ok = false;
				}
				if(ok){
					if(j >= 1 && ((k>>(j-1)) & 1)){
						dp[i+1][j-1] += dp[i][j];
					}
					else if(j <= W-2 && ((k >> j) & 1)){
						dp[i+1][j+1] += dp[i][j];
					}
					else{
						dp[i+1][j] += dp[i][j];
					}
				}
			}
		}
	}
	cout << dp[H][K-1].x << endl;
	return 0;
}