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;
typedef long long ll;
#define int ll

#define REP(i,n) for(int i=0;i<n;++i)
#define SORT(name) sort(name.begin(), name.end())
#define ZERO(p) memset(p, 0, sizeof(p))
#define MINUS(p) memset(p, -1, sizeof(p))
#if 1
#  define DBG(fmt, ...) printf(fmt, ##__VA_ARGS__)
#else
#  define DBG(fmt, ...)
#endif

const ll LLINF = (1LL<<60);
const int INF = (1LL<<30);
const double DINF = std::numeric_limits<double>::infinity();
const int MOD = 1000000007;
#define MAX_H 110
#define MAX_W 10

// n を size 桁の二進数文字列表現 string にして返す
string Binf(int n, int size) {
    string res = "";
    REP(i, size) {
        if((n >> i) & 1) { res += '1'; }
        else { res += '0'; }
    }
    reverse(res.begin(), res.end());
    return res;
}

ll dp[MAX_H][MAX_W] = {};
ll H, W, K;
vector<ll> all_pat;

// 現在 i 行 j 列にいるときのパターン数
ll dfs(ll i = 0, ll j = 0) {
    if(i >= H) {
        if(j == K) { return 1; }
        return 0;
    }

    ll& ret = dp[i][j];
    if(~ret) { return ret; }
    ret = 0;

    for(auto& pat : all_pat) {
        if(j > 0 && ((pat >> (j-1)) & 1) ) {
            // 左に移動するパターン
            ret += dfs(i+1, j-1);
            ret %= MOD;
        } else if(j < W-1 && ((pat >> (j)) & 1) ) {
            // 右に移動するパターン
            ret += dfs(i+1, j+1);
            ret %= MOD;
        } else {
            ret += dfs(i+1, j);
            ret %= MOD;
        }
    }

    return ret;
}

signed main()
{
    cin >> H >> W >> K;
    MINUS(dp);
    K--;

    // all_pat の作成
    for(int pat = 0; pat < (1 << (W-1)); ++pat) {
        // 隣り合うビットが 1 のものはダメ
        bool isOk = true;
        for(int i = 0; i < W-2; ++i) {
            if( ((pat >> i) & 1) && ((pat >> (i+1)) & 1) ) {
                isOk = false;
                break;
            }
        }
        if(isOk) { all_pat.push_back(pat); }
    }

    printf("%lld\n", dfs());
    return 0;
}