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(decltype(n) i = 0; i < n; i++)

#define ALL(c)   c.begin(), c.end()
#define SORT(c)  std::sort(ALL(c))
#define RSORT(c) std::sort(ALL(c), std::greater<decltype(c)::value_type>())

using namespace std;
using ll = long long;

const int MOD  = (int)1e9 + 7;
const int INF  = (int)1e9 + 1;
const ll  LINF = (ll)1e18 + 1;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int H, W, K;
    std::cin >> H >> W >> K;

    vector<vector<ll>> dp(H + 1, vector<ll>(W, 0));

    dp[0][0] = 1;
    for(int hi = 1; hi < H + 1; hi++) {
        REP(wi, W) {
            // bit全探索
            for(int i = 0; i < (1 << (W - 1)); i++) {
                [&]() {
                    bool prev_is_assert = false;
                    for(int j = 0; j < W - 1; j++) {
                        if(i & (1 << j)) {
                            if(prev_is_assert) {
                                return;
                            }
                            else {
                                prev_is_assert = true;
                            }
                        }
                        else {
                            prev_is_assert = false;
                        }
                    }

                    bool l = i & (1 << (wi - 1));
                    bool r = i & (1 << wi);
                    if(l) {
                        dp[hi][wi] += dp[hi - 1][wi - 1];
                        dp[hi][wi] %= MOD;
                    }
                    else if(r) {
                        dp[hi][wi] += dp[hi - 1][wi + 1];
                        dp[hi][wi] %= MOD;
                    }
                    else {
                        dp[hi][wi] += dp[hi - 1][wi];
                        dp[hi][wi] %= MOD;
                    }
                }();
            }
        }
    }

    std::cout << dp[H][K - 1] << std::endl;

    return 0;
}