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"
using namespace std;
using ll = long long;
#define rep(i, begin, n) for (int i = begin; i < n; i++)
#define repe(i, begin, n) for (int i = begin; i <= n; i++)
#define repr(i, begin, n) for (int i = begin; i > begin - n; i--)
#define repre(i, begin, end) for (int i = begin; i >= end; i--)

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 inf = 1000000007;
const int MOD = 1000000007;
const long long INF = 1000000000000000007;

// -------------------------------------------------------

int H, W, K;
ll dp[110][20];

int main()
{
    cin >> H >> W >> K;
    dp[0][1] = 1;
    for (int i = 0; i <= H; i++)
    {
        for (int j = 1; j <= W; j++)
        {
            for (int bit = 0; bit < (1 << (W - 1)); bit++)
            {
                bool ok = true;
                for (int k = 0; k < W - 1; k++)
                {
                    if ((bit & (1 << k)) && (bit & (1 << (k + 1))))
                    {
                        ok = false;
                    }
                }
                if (ok)
                {
                    int nj = j;
                    if (bit & (1 << (W - j)))
                    {
                        nj = j - 1;
                    }
                    else if (bit & (1 << (W - j - 1)))
                    {
                        nj = j + 1;
                    }
                    dp[i + 1][nj] += dp[i][j];
                    dp[i + 1][nj] %= MOD;
                }
            }
        }
    }
    cout << dp[H][K] << endl;
}