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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
/* typedef */
typedef long long ll;
typedef pair<int, int> pii;
/* constant */
const int INF = 1 << 30;
const ll LINF = 1LL << 50;
const int NIL = -1;
const int MAX = 10000;
const int mod = 1000000007;
const double pi = 3.141592653589;
/* struct for mod Calculation*/
struct mint {
    ll x;
    mint(ll x = 0):x((x % mod + mod) % mod) {}
    mint operator-() const { return mint(-x); }
    mint& operator+=(const mint a) {
        if ((x += a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator-=(const mint a) {
        if ((x += mod - a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator*=(const mint a) {
        (x *= a.x) %= mod;
        return *this;
    }
    mint operator+(const mint a) const {
        mint res(*this);
        return res += a;
    }
    mint operator-(const mint a) const {
        mint res(*this);
        return res -= a;
    }
    mint operator*(const mint a) const {
        mint res(*this);
        return res *= a;
    }
    mint pow(ll t) const {
        if (!t) return 1;
        mint a = pow(t >> 1);
        a *= a;
        if (t & 1) a *= *this;
        return a;
    }

    // for prime mod
    mint inv() const {
        return pow(mod - 2);
    }
    mint& operator/=(const mint a) {
        return (*this) *= a.inv();
    }
    mint operator/(const mint a) const {
        mint res(*this);
        return res /= a;
    }
};
/* global variables */
int H, W, K;
vector<vector<ll> > lines;
vector<vector<mint> > dp;
/* function */
void makeLines() {
    int upper = 1 << (W - 1);
    vector<ll> tmp(W - 1);
    for (int b = 0; b < upper; b++) {
        for (int i = 0; i < W - 1; i++) tmp[i] = b >> i & 1;
        bool canLine = true;
        for (int i = 0; i < W - 2; i++) {
            if (tmp[i] + tmp [i + 1] == 2) {
                canLine = false;
                break;
            }
        }
        if (canLine) lines.push_back(tmp);
    }
}
void calcDPTable() {
    dp.assign(H + 1, vector<mint>(W));
    dp[0][0] = 1;
    for (int i = 0; i < H; i++) {
        for (vector<ll> line : lines) {
            for (int j = 0; j < W; j++) {
                if (j > 0 && line[j - 1]) { dp[i + 1][j - 1] += dp[i][j]; }
                else if (j < W && line[j]) { dp[i + 1][j + 1] += dp[i][j]; }
                else { dp[i + 1][j] += dp[i][j]; }
            }
        }
    }
}

void calcAns() { cout << dp[H][K - 1].x << '\n'; }
/* main */
int main(){
    cin >> H >> W >> K;
    if (W == 1) { cout << 1 << '\n'; return 0; }
    makeLines();
    calcDPTable();
    calcAns();
}