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 _overload3(_1, _2, _3, name, ...) name
#define _rep(i, n) repi(i, 0, n)
#define repi(i, a, b) for (int i = (int)(a); i < (int)(b); ++i)
#define rep(...) _overload3(__VA_ARGS__, repi, _rep, )(__VA_ARGS__)
#define ll long long
#define lld long double
#define ALL(x) x.begin(), x.end()
using namespace std;

const int IINF = 1 << 30;
const ll INF = 1ll << 60;
const ll MOD = 1e9 + 7;
int h, w, k;

bool check(int num)
{
    bool prevone = false;
    while (num != 0)
    {
        int cur = num % 2;
        if (cur == 1)
        {
            if (prevone == true)
                return false;
            prevone = true;
        }
        else
        {
            prevone = false;
        }
        num /= 2;
    }
    return true;
}

int main(int argc, char const *argv[])
{
    cin >> h >> w >> k;
    vector<vector<ll>> dp(h + 1, vector<ll>(w, 0));
    dp[0][0] = 1;
    rep(i, 1, h + 1)
    {
        rep(pattern, 1 << (w - 1))
        {
            if (check(pattern) == false)
                continue;
            vector<int> order(w);
            rep(j, w) order[j] = j;
            rep(j, w - 1)
            {
                if (((pattern >> j) & 1) == 1)
                {
                    swap(order[j], order[j + 1]);
                }
            }
            rep(j, w)
            {
                dp[i][j] += dp[i - 1][order[j]];
                dp[i][j] %= MOD;
            }
        }
    }
    /*
    rep(i, h + 1)
    {
        for (auto x : dp[i])
            cerr << x << " ";
        cerr << endl;
    }*/
    cout << dp[h][k - 1] << endl;
    return 0;
}