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>

long long f(std::vector<std::vector<long long>>& ss, long long i) {
    if (i <= 0) return 1;
    else return ss[i][0] + ss[i][1];
}

int main() {
    long long h, w, k;
    const auto mod = 1000000007ll;
    std::cin >> h >> w >> k;
    if (w == 1) {
        std::cout << 1 << std::endl;
        return 0;
    }
    std::vector<std::vector<long long>> ss(w + 10, std::vector<long long>(2));
    ss[1][0] = 1;
    ss[1][1] = 1;
    for (int i = 2; i < w; i++) {
        ss[i][0] = (ss[i - 1][0] + ss[i - 1][1]) % mod;
        ss[i][1] = ss[i - 1][0];
    }

    std::vector<std::vector<long long>> sw(w + 10, std::vector<long long>(2));
    for (int i = 0; i < w; i++) {
        sw[i][0] = f(ss, i) * f(ss, w - 2 - i) % mod;
        sw[i][1] = f(ss, i - 1) * f(ss, w - 3 - i) % mod;
    }

    // for (int i = 0; i < w; i++) std::cout << sw[i][0] << " " << sw[i][1] << std::endl;

    std::vector<long long> dp(w + 10);
    auto tmp(dp);

    dp[0] = 1;
    auto t = (sw[0][0] + sw[0][1]) % mod;

    for (int i = 1; i <= h; i++) {
        tmp[0] = (sw[0][0] * dp[0] % mod + sw[0][1] * dp[1] % mod) % mod;
        tmp[w - 1] = (sw[w - 2][0] * dp[w - 1] % mod + sw[w - 2][1] * dp[w - 2] % mod) % mod;
        for (int j = 1; j < w - 1; j++) {
            tmp[j] = sw[j - 1][1] * dp[j - 1] % mod;
            tmp[j] += sw[j][1] * dp[j + 1] % mod;
            tmp[j] += (mod + t - sw[j - 1][1] - sw[j][1]) * dp[j] % mod;
            tmp[j] %= mod;
        }
        tmp.swap(dp);
    }

    std::cout << dp[k - 1] << std::endl;

    return 0;
}