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 ALL(g) (g).begin(),(g).end()
#define REP(i, x, n) for(int i = x; i < n; i++)
#define rep(i,n) REP(i,0,n)
#define RREP(i, x, n) for(int i = x; i >= n; i--)
#define rrep(i, n) RREP(i,n,0)
#define pb push_back
#pragma GCC optimize ("-O3")

using namespace std;

#define DEBUG_IS_VALID

#ifdef DEBUG_IS_VALID
#define DEB 1 
#else
#define DEB 0
#endif
#define DUMPOUT cout
#define dump(...) if(DEB) DUMPOUT<<"  "<<#__VA_ARGS__<<" :["<<__LINE__<<":"<<__FUNCTION__<<"]"<<endl<<"    "; if(DEB) dump_func(__VA_ARGS__)
template<typename T1,typename T2>ostream& operator << (ostream& os, pair<T1,T2> p){cout << "(" << p.first << ", " << p.second << ")"; return os;}
template<typename T>ostream& operator << (ostream& os, vector<T>& vec) { os << "{"; for (int i = 0; i<vec.size(); i++) os << vec[i] << (i + 1 == vec.size() ? "" : ", "); os << "}"; return os; }
template<typename T>ostream& operator << (ostream& os, set<T>& st){for(auto itr = st.begin(); itr != st.end(); itr++) cout << *itr << (next(itr)!=st.end() ? ", " : ""); cout << "}"; return os;}
template<typename T1,typename T2>ostream& operator << (ostream& os, map<T1,T2> mp){cout << "{"; for(auto itr = mp.begin(); itr != mp.end(); itr++) cout << "(" << (itr->first) << ", " << (itr->second) << ")" << (next(itr)!=mp.end() ? "," : ""); cout << "}"; return os; }

void dump_func(){DUMPOUT << endl;}
template <class Head, class... Tail>void dump_func(Head&& head, Tail&&... tail){ DUMPOUT << head; if (sizeof...(Tail) == 0) { DUMPOUT << " "; } else { DUMPOUT << ", "; } dump_func(std::move(tail)...);}
template<class T> inline bool chmax(T& a,T const& b){if(a>=b) return false; a=b; return true;}
template<class T> inline bool chmin(T& a,T const& b){if(a<=b) return false; a=b; return true;}

using ll = long long;
using P = pair<int,int>;
using Pl = pair<ll,ll>;
using vi = vector<int>;
using vvi = vector<vi>;
using vl = vector<ll>;
using vvl = vector<vl>;

const int mod=1e9+7,INF=1<<29;
const double EPS=1e-12,PI=3.1415926535897932384626;
const ll lmod = 1e9+7,LINF=1LL<<59; 

template<int MOD> struct ModInt {
    static const int Mod = MOD; unsigned x; ModInt() : x(0) { }
    ModInt(signed sig) { x = sig < 0 ? sig % MOD + MOD : sig % MOD; }
    ModInt(signed long long sig) { x = sig < 0 ? sig % MOD + MOD : sig % MOD; }
    int get() const { return (int)x; }
    ModInt &operator+=(ModInt that) { if ((x += that.x) >= MOD) x -= MOD; return *this; }
    ModInt &operator-=(ModInt that) { if ((x += MOD - that.x) >= MOD) x -= MOD; return *this; }
    ModInt &operator*=(ModInt that) { x = (unsigned long long)x * that.x % MOD; return *this; }
    ModInt &operator/=(ModInt that) { return *this *= that.inverse(); }
    ModInt operator+(ModInt that) const { return ModInt(*this) += that; }
    ModInt operator-(ModInt that) const { return ModInt(*this) -= that; }
    ModInt operator*(ModInt that) const { return ModInt(*this) *= that; }
    ModInt operator/(ModInt that) const { return ModInt(*this) /= that; }
    ModInt inverse() const { long long a = x, b = MOD, u = 1, v = 0;
        while (b) { long long t = a / b; a -= t * b; std::swap(a, b); u -= t * v; std::swap(u, v); }
        return ModInt(u); }
    bool operator==(ModInt that) const { return x == that.x; }
    bool operator!=(ModInt that) const { return x != that.x; }
    ModInt operator-() const { ModInt t; t.x = x == 0 ? 0 : Mod - x; return t; }
};
template<int MOD> ostream& operator<<(ostream& st, const ModInt<MOD> a) { st << a.get(); return st; };
template<int MOD> ModInt<MOD> operator^(ModInt<MOD> a, unsigned long long k) {
    ModInt<MOD> r = 1; while (k) { if (k & 1) r *= a; a *= a; k >>= 1; } return r; }
template<typename T, int FAC_MAX> struct Comb { vector<T> fac, ifac;
    Comb(){fac.resize(FAC_MAX,1);ifac.resize(FAC_MAX,1); for(int i = 1; i < FAC_MAX; i++)fac[i]=fac[i-1]*i;
        ifac[FAC_MAX-1]=T(1)/fac[FAC_MAX-1];for(int i = FAC_MAX-2; i >= 1; i--)ifac[i]=ifac[i+1]*T(i+1);}
    T aPb(int a, int b) { if (b < 0 || a < b) return T(0); return fac[a] * ifac[a - b]; }
    T aCb(int a, int b) { if (b < 0 || a < b) return T(0); return fac[a] * ifac[a - b] * ifac[b]; }
    T nHk(int n, int k) { if (n == 0 && k == 0) return T(1); if (n <= 0 || k < 0) return 0;
        return aCb(n + k - 1, k); }}; // nHk = (n+k-1)Ck : n is separator
typedef ModInt<1000000007> mint;

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  int h,w,k; cin >> h >> w >> k ;
  using vm = vector<mint>;
  vector<vm> dp(h+1,vm(w));
  dp[0][0] = mint(1);
  auto chk = [&](int k){
    bool ok = true;
    rep(x,w-2) ok &= !(((k>>x)&1)==1 && ((k>>(x+1))&1)==1);
    return ok;
  };
  rep(i,h){
    rep(k,1<<(w-1)){
      if(!chk(k)) continue;
      rep(x,w){
        if(((k>>x)&1)==1) dp[i+1][x+1] += dp[i][x];
        else if(x>0 && ((k>>(x-1))&1)==1) dp[i+1][x-1] += dp[i][x];
        else dp[i+1][x] += dp[i][x];
      }
    }
  }
  cout << dp[h][k-1] << endl;
  return 0;
}