#include <bits/stdc++.h>

using namespace std;
using ll = long long;

const ll modc = 998244353;
class mint {
    ll x;
public:
    mint(ll x=0) : x((x%modc+modc)%modc) {}
    mint operator-() const { return mint(-x);}
    mint& operator+=(const mint& a) {
        if ((x += a.x) >= modc) x -= modc;
        return *this;
    }
    mint& operator-=(const mint& a) {
        if ((x += modc-a.x) >= modc) x -= modc;
        return *this;
    }
    mint& operator*=(const  mint& a) {
        (x *= a.x) %= modc;
        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;
    }
    mint inv() const {return pow(modc-2);}
    mint& operator/=(const mint& a){ return (*this) *= a.inv();}
    mint operator/(const mint& a) const {
        mint res(*this);
        return res/=a;
    }
    bool operator == (const mint& a) const{ return x == a.x;}
    bool operator != (const mint& a) const{ return x != a.x;}
    friend ostream& operator<<(ostream& os, const mint& m){
        os << m.x;
        return os;
    }
    friend istream& operator>>(istream& ip, mint &m) {
        ll t;
        ip >> t;
        m = mint(t);
        return ip;
    }
    ll val(){ return x;}
};

int main(){
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);

    int N, M, K, L;
    cin >> N >> M >> K;
    L = 1<<N;

    vector dp(M+1, vector<mint>(L));
    for (int i=0; i<L; i++){
        if (__builtin_popcount(i) >= K) dp[1][i] = 1;
    }

    for (int i=2; i<=M; i++){
        for (int j=0; j<L; j++){
            for (int k=0; k<L; k++){
                if (__builtin_popcount(j&k) >= K){
                    dp[i][j] += dp[i-1][k];
                }
            }
        }
    }

    mint ans = 0;
    for (int i=0; i<L; i++) ans += dp[M][i];
    cout << ans << endl;

    return 0;
}