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 <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <string>
#include <sstream>
#include <complex>
#include <vector>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
#include <numeric>
using namespace std;
typedef long long int ll;

#define EPS (1e-7)
#define INF 1e18
#define max(p,q)((p)>(q)?(p):(q))
#define min(p,q)((p)<(q)?(p):(q))
#define PI (acos(-1))

#define REP(i, n) for(ll i = 0; i < (ll)(n); i++)


int main() {

    const ll MOD = 1e9 + 7;
    ll H, W, K;
    cin >> H >> W >> K;
    vector<vector<ll>> v(H+1, vector<ll>(W, 0));
    v[0][0] = 1;

    vector<vector<ll>> a(W, vector<ll>(W, 0));
    vector<ll> tmp(W);
    bool is_suit;
    REP(i, 1<<(W-1)){
        is_suit = true;
        REP(j, W-1) tmp[j] = (i>>j) & 1;
        REP(j, W-2) if (tmp[j] == 1 && tmp[j+1] == 1) is_suit = false;
        if (!is_suit) continue;
        REP(j, W){
            if (j != W-1 && tmp[j] == 1) a[j][j+1]++;
            else if (j != 0 && tmp[j-1] == 1) a[j][j-1]++;
            else a[j][j]++;
        }
    }

    /*
    REP(i, W) {
        REP(j, W) cout << a[i][j];
        cout << endl;
    }
    
*/
   REP(i, H)REP(j, W){
       v[i+1][j] = (v[i+1][j] + (v[i][j] * a[j][j]) % MOD) % MOD;
       if (j != W-1) v[i+1][j+1] = (v[i+1][j+1] + (v[i][j] * a[j][j+1] % MOD)) % MOD;
       if (j != 0) v[i+1][j-1] = (v[i+1][j-1] + (v[i][j] * a[j][j-1] % MOD)) % MOD;
   }
   /*
    REP(i, H+1) {
        REP(j, W) cout << v[i][j];
        cout << endl;
    }
*/
   cout << v[H][K-1] << endl;

    return 0;
}