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>
#define REP(i,n) for(int i=0,i##_len=(n);i<i##_len;++i)
#define RREP(i,n) for(int i=int(n)-1;i>=0;--i)
#define rep(i,a,b) for(int i=int(a);i<int(b);++i)
#define rrep(i,a,b) for(int i=int(a)-1;i>=int(b);--i)
#define All(x) (x).begin(),(x).end()
#define rAll(x) (x).rbegin(),(x).rend()
#define ITR(i,x) for(auto i=(x).begin();i!=(x).end();++i)
using namespace std;
using Graph = vector<vector<int>>;

typedef long long ll;
typedef pair<ll, ll> P;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<vl> vvl;
constexpr ll mod = 1e9+7;
constexpr double eps = 1e-9;

void solve();
P bisearch(ll l,ll r,function<bool(ll)> f){
    while((l+1)<r){
        ll m=(l+r)/2;
        if(f(m)) r=m;
        else l=m;
    }
    return P(l,r);
}
ll SQRT(ll n){if(n==1) return 1;return bisearch(0,n,[n](ll x){return x>n/x;}).first;}
ll roundup(ll n,ll div){
    if(n>0) return (n-1)/div+1;
    else return n/div;    
}
bool square(ll a){ll n=SQRT(a);return a==n*n;}
ll npow(ll x, ll n){
    ll ans = 1;
    while(n != 0){
        if(n&1) ans = ans*x;
        x = x*x;
        n = n >> 1;
    }
    return ans;
}
ll mpow(ll x, ll n){ //x^n(mod) ←普通にpow(x,n)では溢れてしまうため，随時mod計算 2分累乗法だから早い
    ll ans = 1;
    while(n != 0){
        if(n&1) ans = ans*x % mod;
        x = x*x % mod;
        n = n >> 1;
    }
    return ans;
}
ll inv_mod(ll a){return mpow(a,mod-2);}
int digitsum(ll N,int a){
    if(N==0) return 0;
    int ret=0;
    ret+=digitsum(N/a,a)+N%a;
    return ret;
}
ll gcd(ll x,ll y){return y ? gcd(y,x%y) : x;};//xとyの最大公約数
ll lcm(ll x,ll y){return x/gcd(x,y)*y;}//xとyの最小公倍数
void YN(bool flg){std::cout<<(flg?"YES":"NO")<<endl;}
void Yn(bool flg){std::cout<<(flg?"Yes":"No")<<endl;}
void yn(bool flg){std::cout<<(flg?"yes":"no")<<endl;}
P splitint(string n,int a){
    int Len=n.length();
    if(a<0||Len<a) return {-1,-1};
    string left,right;
    if(a!=0) left=n.substr(0,a);
    if(a!=Len) right=n.substr(a);
    return P(stoll(left),stoll(right));
}
ll manhattan(const P &a,const P &b){return llabs(a.first-b.first)+llabs(a.second-b.second);}
bool kaibun(string s){//O(|S|)
    string  sdash=s;
    reverse(All(s));
    return s==sdash;
}
class Ruiseki{
  private:
    vector<ll> LEFT,RIGHT;
    function<ll(ll,ll)> F;
    int N;
  public:
    Ruiseki(ll INI,const vector<ll> &a,function<ll(ll,ll)> f){
        N=a.size();F=f;
        LEFT.resize(N+1);RIGHT.resize(N+1);
        LEFT[0]=RIGHT[0]=INI;
        REP(i,N){ 
            LEFT[i+1]=F(LEFT[i],a[i]);
            RIGHT[i+1]=F(RIGHT[i],a[N-i-1]);
        }
    }
    ll out(int l,int r){//[l,r)の外の累積
        return F(LEFT[l],RIGHT[N-r]);
    }
};
class mint {
 private:
  ll _num,_mod;
  mint set(ll num){ 
      _num = num ;
      if(_num>=0) _num%=_mod;
      else _num+=(1-(_num+1)/_mod)*_mod; 
      return *this;
  }
  ll imod(ll n){return mpow(n , _mod-2);}
 public:
  mint(){ _num = 0;_mod=mod; }
  mint(ll num){ _mod = mod; _num = (num+(1LL<<25)*mod) % mod; }
  mint(ll num,ll M){ _mod=M;_num=(num+(1LL<<25)*mod)%_mod; }
  mint(const mint &cp){_num=cp._num;_mod=cp._mod;}
  mint operator= (const ll x){ return set(x); }
  mint operator+ (const ll x){ return mint(_num + (x % _mod) , _mod); }
  mint operator- (const ll x){ return mint(_num - (x % _mod), _mod); }
  mint operator* (const ll x){ return mint(_num * (x % _mod) , _mod); }
  mint operator/ (ll x){ return mint(_num * imod(x) , _mod);}
  mint operator+=(const ll x){ return set(_num + (x % _mod)); }
  mint operator-=(const ll x){ return set(_num - (x % _mod)); }
  mint operator*=(const ll x){ return set(_num * (x % _mod)); }
  mint operator/=(ll x){ return set(_num * imod(x));}
  mint operator+ (const mint &x){ return mint(_num + x._num , _mod); }
  mint operator- (const mint &x){ return mint(_num - x._num , _mod);}
  mint operator* (const mint &x){ return mint(_num * x._num , _mod); }
  mint operator/ (mint x){ return mint(_num * imod(x._num) , _mod);}
  mint operator+=(const mint &x){ return set(_num + x._num); }
  mint operator-=(const mint &x){ return set(_num - x._num); }
  mint operator*=(const mint &x){ return set(_num * x._num); }
  mint operator/=(mint x){ return set(_num * imod(x._num));}

  bool operator<(const mint &x){return _num<x._num;}

  friend mint operator+(ll x,const mint &m){return mint(m._num + (x % m._mod) , m._mod);}
  friend mint operator-(ll x,const mint &m){return mint(m._num - (x % m._mod) , m._mod);}
  friend mint operator*(ll x,const mint &m){return mint(m._num * (x % m._mod) , m._mod);}
  friend mint operator/(ll x,mint m){return mint(m.imod(m._num) * x , m._mod);}

  explicit operator ll() { return _num; }
  explicit operator int() { return (int)_num; }

  friend ostream& operator<<(ostream &os, const mint &x){ os << x._num; return os; }
  friend istream& operator>>(istream &is, mint &x){ll val; is>>val; x.set(val); return is;}
};
template<typename T> class MAT{
 private:
    int row,col;
    vector<vector<T>> _A;
    MAT set(vector<vector<T>> A){_A = A ; return *this;}
    
 public:
    MAT(){ }
    MAT(int n,int m){
        if(n<1 || m<0){cout << "err Matrix::Matrix" <<endl;exit(1);}
        row = n;
        col = m?m:n;//m=0のとき単位行列を作る
        REP(i,row){
            vector<T> a(col);
            _A.push_back(a);
            REP(j,col) _A[i][j]=0;
        }
        //  値の初期化
        if(m==0) REP(i,n) _A[i][i]=1.0;
    }
    MAT(const MAT &cp){_A=cp._A;row=cp.row;col=cp.col;}
    T* operator[] (int i){return _A[i].data();}
    MAT operator= (vector<vector<T>> x) {return set(x);}
    MAT operator+ (MAT x) {
        if(row!=x.row || col!=x.col){
            cout << "err Matrix::operator+" <<endl;
            cout << "  not equal matrix size" <<endl;
            exit(0);
        }
        MAT r(row, col);
        REP(i,row) REP(j,col) r[i][j]=_A[i][j]+x[i][j];
        return r;
    }
    MAT operator- (MAT x) {
        if(row!=x.row || col!=x.col){
            cout << "err Matrix::operator-" <<endl;
            cout << "  not equal matrix size" <<endl;
            exit(0);
        }
        MAT r(row, col);
        REP(i,row) REP(j,col) r[i][j]=_A[i][j]-x[i][j];
        return r;
    }
    MAT operator* (MAT x) {
        if(col!=x.row){
            cout << "err Matrix::operator*" <<endl;
            cout << "  not equal matrix size" <<endl;
            exit(0);
        }
        MAT r(row, x.col);
        REP(i,row) REP(j,x.col) REP(k,col) r[i][j]+=_A[i][k]*x[k][j];
        return r;
    }
    MAT operator/ (T a){
        MAT r(row,col);
        REP(i,row) REP(j,col) r[i][j]=_A[i][j]/a;
        return r;
    }
    MAT operator^ (ll n){
        if(row!=col){
            cout << "err Matrix::operator^" <<endl;
            cout << "  not equal matrix size" <<endl;
            exit(0);
        }
        MAT r(row,0),A=*this;
        while(n){
            if(n&1) r *= A;
            A*=A;
            n>>=1;
        }
        return r;
    }
    MAT operator+= (MAT x) {
        if(row!=x.row || col!=x.col){
            cout << "err Matrix::operator+=" <<endl;
            cout << "  not equal matrix size" <<endl;
            exit(0);
        }
        MAT r(row, col);
        REP(i,row) REP(j,col) r[i][j]=_A[i][j]+x[i][j];
        return set(r._A);
    }
    MAT operator-= (MAT x) {
        if(row!=x.row || col!=x.col){
            cout << "err Matrix::operator-=" <<endl;
            cout << "  not equal matrix size" <<endl;
            exit(0);
        }
        MAT r(row, col);
        REP(i,row) REP(j,col) r[i][j]=_A[i][j]-x[i][j];
        return set(r._A);
    }
    MAT operator*= (MAT x) {
        if(col!=x.row){
            cout << "err Matrix::operator*" <<endl;
            cout << "  not equal matrix size" <<endl;
            exit(0);
        }
        MAT r(row, x.col);
        REP(i,row) REP(j,x.col) REP(k,col) r[i][j]+=_A[i][k]*x[k][j];
        return set(r._A);
    }
    MAT operator/=(T a){
        MAT r(row,col);
        REP(i,row) REP(j,col) r[i][j]=_A[i][j]/a;
        return r;
    }

    friend MAT operator* (T n,MAT x){
        MAT r(x.row,x.col);
        REP(i,x.row) REP(j,x.col) r[i][j]=n*x[i][j];
        return r;
    }
    friend MAT operator* (MAT x,T n){
        MAT r(x.row,x.col);
        REP(i,x.row) REP(j,x.col) r[i][j]=n*x[i][j];
        return r;
    }
    explicit operator vector<vector<T>>(){return _A;}
    friend ostream &operator<<(ostream &os,const MAT &x){ REP(i,x.row) REP(j,x.col) os<<x._A[i][j]<<" \n"[j==x.col-1]; return os;}
    friend istream &operator>>(istream &is,MAT &x){REP(i,x.row) REP(j,x.col) is>>x._A[i][j];return is;}
    int size_row(){return row;}
    int size_col(){return col;}
    MAT transpose(){
        MAT r(col,row);
        REP(i,col) REP(j,row) r[i][j]=_A[j][i];
        return r;
    }
    MAT inverse(){
        T buf;
        MAT<T> inv_a(row,0);
        vector<vector<T>> a=_A;
        //掃き出し法
        REP(i,row){
            buf=1/a[i][i];
            REP(j,row){
                a[i][j]*=buf;
                inv_a[i][j]*=buf;
            }
            REP(j,row){
                if(i!=j){
                    buf=a[j][i];
                    REP(k,row){
                        a[j][k]-=a[i][k]*buf;
                       inv_a[j][k]-=inv_a[i][k]*buf;
                    }
                }
            }
        }
        return inv_a;
    }
    // O( n^3 ).
    int rank() {
        vector<vector<T>> A=_A;
        const int n = row, m = col;
        int r = 0;
        for(int i = 0; r < n && i < m; ++i) {
            int pivot = r;
            for(int j = r+1; j < n; ++j) if(abs(A[j][i]) > abs(A[pivot][i])) pivot = j;
            swap(A[pivot], A[r]);
            if(abs(A[r][i]) < eps) continue;
            for (int k = m-1; k >= i; --k) A[r][k] /= A[r][i];
            rep(j,r+1,n) rep(k,i,m) A[j][k] -= A[r][k] * A[j][i];
            ++r;
        }
        return r;
    }
};
class UnionFind{//UnionFind木
 private:
    vector<int> Parent;
    vector<ll> diff_weight;
 public:
    UnionFind(int N){
        Parent.resize(N,-1);
        diff_weight.resize(N,0LL);
    }

    int root(int A){
        if(Parent[A]<0) return A;
        else{ 
            int r = root(Parent[A]);
            diff_weight[A] += diff_weight[Parent[A]]; // 累積和をとる
            return Parent[A]=r;
        }
    }
    bool issame(int A,int B){
        return root(A)==root(B);
    }
    ll weight(int x) {
        root(x); // 経路圧縮
        return diff_weight[x];
    }
    ll diff(int x, int y) {
        return weight(y) - weight(x);
    }
    int size(int A){
        return -Parent[root(A)];
    }

    bool connect(int A,int B){
        A=root(A); B=root(B);
        if(A==B) return false;
        if(size(A)<size(B)) swap(A,B);
        Parent[A]+=Parent[B];
        Parent[B]=A;
        return true;
    }
    bool merge(int A, int B, ll w) {
        // x と y それぞれについて、 root との重み差分を補正
        w += weight(A); w -= weight(B); 
        A=root(A); B=root(B);
        if(A==B) return false;
        if(size(A)<size(B)) swap(A,B),w=-w;
        Parent[A]+=Parent[B];
        Parent[B]=A;
        // x が y の親になるので、x と y の差分を diff_weight[y] に記録
        diff_weight[B] = w; 
        return true;
    }
};
class Factorial{//階乗とその逆元を求めて計算に利用するクラス
 private:
    vector<ll> fac;
    vector<ll> ifac;
 public:
    
    Factorial(ll N){
        fac.push_back(1);
        REP(i,N) fac.push_back(fac[i]*(i+1)%mod);
        ifac.resize(N+1);
        ifac[N]=inv_mod(fac[N]);
        REP(i,N) ifac[N-1-i]=(ifac[N-i]*(N-i))%mod;
    }

    ll fact(ll a){return fac[a];}
    ll ifact(ll a){return ifac[a];}

    ll cmb(ll a,ll b){
        if(a==0&&b==0) return 1;
        if(a<b||a<0||b<0) return 0;
        ll tmp =ifact(a-b)*ifact(b)%mod;
        return tmp*fac[a]%mod;
    }
    ll per(ll a,ll b){
        if(a==0&&b==0) return 1;
        if(a<b||a<0||b<0) return 0;
        return fac[a]*ifac[a-b]%mod;
    }
};
class SOSU{
 private:
    vector<ll> Prime_Number;
 public:
    SOSU(ll N){
        vector<bool> isp(N+1,true);
        isp[0]=isp[1]=false;
        rep(i,2,N+1){
            if(isp[i]){
                Prime_Number.push_back(i);
                for(int j=2*i;j<=N;j+=i) isp[j]=false;
            }
        }
    }
    ll operator[](int i){return Prime_Number[i];}
    int size(){return Prime_Number.size();}
    ll back(){return Prime_Number.back();}
    bool isPrime(int q){return binary_search(All(Prime_Number),q);}
};
class Divisor{//素因数分解をしてから約数列挙、分解結果はＰ(底,指数)でpfacにまとめている
  private:
    vector<ll> F;
    vector<P> pfactorize;
  public:
    Divisor(ll N){
        for(ll i = 1; i * i <= N; i++) {
            if(N % i == 0) {
                F.push_back(i);
                if(i * i != N) F.push_back(N / i);
            }
        }
        sort(begin(F), end(F));
		SOSU p(SQRT(N)+1);
    	REP(i,p.size()){
    		pfactorize.push_back(P(p[i],0));
    		while(N%p[i]==0){
    			N/=p[i];
    			pfactorize.back().second++;
    		}
            if(pfactorize.size()==0) pfactorize.pop_back();
    	}
    	if(N>1) pfactorize.push_back(P(N,1));
    }
    int size(){return F.size();}
    vector<P> pfac(){return pfactorize;}
    ll operator[](int k){return F[k];}
};
struct Solutions{
    ll napsack(int kinds,int MAX_W,const vl &weight,const vl &cost){
        vector<vector<ll>> dp(kinds+1,vector<ll>(MAX_W+1,0));
        REP(i,kinds) REP(j,MAX_W+1){
            if(j<weight[i]) dp[i+1][j]=dp[i][j];
            else dp[i+1][j]=max(dp[i][j],dp[i][j-weight[i]]+cost[i]);
        }
        return dp[kinds][MAX_W];
    }
    ll cost_based_napsack(int kinds,int MAX_W,const vl &weight,const vl &cost){
        int MAX_V=0;
        REP(i,kinds) MAX_V=max(MAX_V,(int)cost[i]);
        vvl dp(kinds+1,vl(kinds*MAX_V+1,1LL<<58));
        dp[0][0] = 0;
        REP(i,kinds) REP(j,kinds*MAX_V+1){
            if(j<cost[i]) dp[i+1][j]=dp[i][j];
            else dp[i+1][j] = min(dp[i][j],dp[i][j-cost[i]]+weight[i]);
        }
        int res=0;
        REP(i,kinds*MAX_V+1) if(dp[kinds][i]<=MAX_W) res=i;
        return res;
    }
    ll unlimited_napsack(int kinds,int MAX_W,const vl &weight,const vl &cost){
        vector<vector<ll>> dp(kinds+1,vector<ll>(MAX_W+1,0));
        REP(i,kinds) REP(j,MAX_W+1){
            if(j<weight[i]) dp[i+1][j]=dp[i][j];
            else dp[i+1][j]=max(dp[i][j],dp[i+1][j-weight[i]]+cost[i]);
        }
        return dp[kinds][MAX_W];
    }
    ll huge_napsack(int kinds,ll MAX_W,const vl &weight,const vl &cost){
        int n2=kinds/2;
        vector<P> ps(1<<(kinds/2));
        REP(i,1<<n2){
            ll sw=0,sv=0;
            REP(j,n2){
                if(i>>j&1){
                    sw += weight[j];
                    sv += cost[j];
                }
            }
            ps[i] = P(sw,sv);
        }
        sort(ps.begin(),ps.begin() + (1<<n2) );
        int m=1;
        rep(i,1,1<<n2){
            if(ps[m-1].second<ps[i].second) ps[m++] = ps[i];
        }

        ll res=0;
        REP(i,1<<(kinds-n2)){
            ll sw=0,sv=0;
            REP(j,kinds-n2){
                if((i>>j)&1){
                    sw += weight[n2+j];
                    sv += cost[n2+j];
                }
            }
            if(sw<=MAX_W){
                ll tv = (lower_bound(ps.begin(),ps.begin()+m,P(MAX_W-sw,LLONG_MAX))-1)->second;
                res = max(res,sv+tv);
            }
        }
        return res;
    }
    ll choose_MonN(int N,int M,const vl &cost){
        vvl dp(N+1,vl(M+1,-1LL<<58));
        REP(i,N+1) dp[i][0]=0;
        REP(i,N) REP(j,M){
            if(j>i) break;
            dp[i+1][j+1]=max(dp[i][j+1],dp[i][j]+cost[i]);
        }
        return dp[N][M];
    }
    ll Partition_Func(int n,int k){
        vector<vector<ll>> dp(k+1,vector<ll> (n+1,0));
        dp[0][0]=1;
        rep(i,1,k+1) REP(j,n+1){
            if(j-i>=0) dp[i][j]=(dp[i][j-i]+dp[i-1][j]);
            else dp[i][j]=dp[i-1][j];
        }
        return dp[k][n];
    }
    int LCS(string s,string t){
        int n=s.length(),m=s.length();
        vector<vector<int>> dp(n+1,vector<int>(m+1));
        REP(i,n) REP(j,m){
            if (s[i] == t[j]) dp[i+1][j+1] = dp[i][j] + 1;
            else dp[i+1][j+1] = max(dp[i][j+1], dp[i+1][j]);
        }
        return dp[n][m];
    }
    int LIS(const vector<ll> a){
        int n=a.size();
        ll INF=1<<28;
        vector<ll> dp(n+1,INF);
        REP(i,n) *lower_bound(All(dp),a[i])=a[i];
        int k=lower_bound(All(dp),INF)-dp.begin();
        return k;
    }
    
    int max_flow(int s,int t,const vector<vector<P>> &g){
        struct edge_{int to,cap, rev;};
        vector<bool> used(g.size(),false);
        vector<vector<edge_>> G(g.size());
        REP(i,g.size()) REP(j,g[i].size()){
            int from = i, to = g[i][j].second;
            int cap = g[i][j].first;
            G[from].push_back((edge_){to,cap,(int)G[to].size()});
            G[to].push_back((edge_){from,0,(int)G[from].size()-1});
        }
        auto dfs = [&](auto&& f,int v,int t,int fl)->int{
            if(v==t) return fl;
            used[v] = true;
            REP(i,G[v].size()){
                edge_ &e = G[v][i];
                if(!used[e.to] && e.cap>0){
                    int d = f(f, e.to,t,min(fl,e.cap));
                    if(d>0){
                        e.cap -= d;
                        G[e.to][e.rev].cap += d;
                        return d;
                    }
                }
            }
            return 0;
        };
        int flow=0;
        while(1){
            used.assign(used.size(),false);
            int f = dfs(dfs,s,t,INT_MAX);
            if(f==0) return flow;
            flow += f;
        }
    }
    int bipartite_matching_calculate(const Graph &g){
        int V = g.size();
        vi match(V,-1);
        vector<bool> used(V,false);
        auto dfs = [&](auto&& f,int v)->bool{
            used[v]=true;
            REP(i,g[v].size()){
                int u = g[v][i], w = match[u];
                if(w<0 || !used[w] && f(f,w)){
                    match[v]=u;
                    match[u]=v;
                    return true;
                }
            }
            return false;
        };
        int res=0;
        REP(v,V){
            if(match[v] < 0){
                used.assign(V,false);
                if(dfs(dfs,v)) res++;
            }
        }
        return res;
    }
    int bipartite_matching(int l,int r,function<bool(int,int)> F){
        int V = l+r;
        Graph G(V);
        REP(i,l) REP(j,r) if(F(i,j)){
            G[i].push_back(l+j);
            G[l+j].push_back(i);
        }
        return bipartite_matching_calculate(G);
    }
};
struct compress{
    map<ll,int> zip;
    vector<int> unzip;
    compress(vector<ll> x){
        sort(All(x));
        x.erase(unique(All(x)),x.end());
        REP(i,x.size()){
            zip[x[i]]=i;
            unzip[i]=x[i];
        }
    }
    int operator[](int k){return zip[unzip[k]];}
};
struct edge{
    int from;int to;ll cost;
    void push(int a,int b,int c){
        from=a;to=b;cost=c;
    }
    bool operator<(const edge& y) const{
        if(cost!=y.cost) return cost<y.cost;
        else if(to!=y.to) return to<y.to;
        else return from<y.from;}
    bool operator>(const edge& y) const{
        if(cost!=y.cost) return cost>y.cost;
        else if(to!=y.to) return to>y.to;
        else return from>y.from;}
    bool operator==(const edge& y) const{return *this==y;}
};
class lca {
  public:
    const int n = 0;
    const int log2_n = 0;
    std::vector<std::vector<int>> parent;
    std::vector<int> depth;

    lca() {}

    lca(const Graph &g, int root)
        : n(g.size()), log2_n(log2(n) + 1), parent(log2_n, std::vector<int>(n)), depth(n) {
        dfs(g, root, -1, 0);
        for (int k = 0; k + 1 < log2_n; k++) {
            for (int v = 0; v < (int)g.size(); v++) {
                if (parent[k][v] < 0)
                    parent[k + 1][v] = -1;
                else
                    parent[k + 1][v] = parent[k][parent[k][v]];
            }
        }
    }

    void dfs(const Graph &g, int v, int p, int d) {
        parent[0][v] = p;
        depth[v] = d;
        REP(j,g[v].size()) {
            if (g[v][j] != p) dfs(g, g[v][j], v, d + 1);
        }
    }

    int get(int u, int v) {
        if (depth[u] > depth[v]) std::swap(u, v);
        for (int k = 0; k < log2_n; k++) {
            if ((depth[v] - depth[u]) >> k & 1) {
                v = parent[k][v];
            }
        }
        if (u == v) return u;
        for (int k = log2_n - 1; k >= 0; k--) {
            if (parent[k][u] != parent[k][v]) {
                u = parent[k][u];
                v = parent[k][v];
            }
        }
        return parent[0][u];
    }
};
template <typename T,typename E> //SegmentTree((要素数) int n_,(演算) F f,(更新) G g,(初期値) T d1)
struct SegmentTree{
  typedef function<T(T,T)> F;
  typedef function<T(T,E)> G;
  int n;
  F f;
  G g;
  T d1;
  E d0;
  vector<T> dat;
  SegmentTree(){};
  SegmentTree(int n_,F f,G g,T d1,
	      vector<T> v=vector<T>()):
    f(f),g(g),d1(d1){
    init(n_);
    if(n_==(int)v.size()) build(n_,v);
  }
  void init(int n_){
    n=1;
    while(n<n_) n*=2;
    dat.clear();
    dat.resize(2*n-1,d1);
  }
  void build(int n_, vector<T> v){
    for(int i=0;i<n_;i++) dat[i+n-1]=v[i];
    for(int i=n-2;i>=0;i--)
      dat[i]=f(dat[i*2+1],dat[i*2+2]);
  }
  void update(int k,E a){
    k+=n-1;
    dat[k]=g(dat[k],a);
    while(k>0){
      k=(k-1)/2;
      dat[k]=f(dat[k*2+1],dat[k*2+2]);
    }
  }
  inline T query(int a,int b){
    T vl=d1,vr=d1;
    for(int l=a+n,r=b+n;l<r;l>>=1,r>>=1) {
      if(l&1) vl=f(vl,dat[(l++)-1]);
      if(r&1) vr=f(dat[(--r)-1],vr);
    }
    return f(vl,vr);
  }
  
};

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    solve();
    return 0;
}

void solve(){
    ll H,W,K;cin>>H>>W>>K;
    vector<vector<mint>> dp(H+1,vector<mint>(W,0));
    dp[0][0]=1;
    K--;
    if(W==1){
        cout<<1<<endl;
        return;
    }
    REP(i,H){ 
        REP(j,1<<(W-1)){
            bool ok=true;
            REP(k,W) if(j>>k&1&&j>>(k+1)&1) ok=false;
            if(!ok) continue; 
            REP(k,W-1){
                if(j>>k&1){
                    dp[i+1][k+1]+=dp[i][k];
                    dp[i+1][k]+=dp[i][k+1];
                }else if(k==0||!j>>(k-1)&1){
                    dp[i+1][k]+=dp[i][k];
                }
            }
        }
        //cout<<dp[i][K]<<endl;
    }
    cout<<dp[H][K]<<endl;
}