#include <bits/stdc++.h>
#include <atcoder/modint>

using namespace std;
using namespace atcoder;
using ll = long long;
using mint = modint998244353;

template<typename T>
using matrix = vector<vector<T>>;

template<typename T>
matrix<T> dot(matrix<T> &a, matrix<T> &b){
    int N = a.size();
    matrix<T> c(N, vector<T>(N));
    for (int i=0; i<N; i++){
        for (int k=0; k<N; k++){
            for (int j=0; j<N; j++) c[i][j] += (a[i][k] * b[k][j]);
        }
    }
    return c;
}

template<typename T>
matrix<T> pow(matrix<T> a, ll N){
    int M = a.size();
    matrix<T> b(M, vector<T>(M));
    for (int i=0; i<M; i++) b[i][i] = 1;
    while(N){
        if (N % 2 == 1) b = dot(b, a);
        N >>= 1;
        a = dot(a, a);
    }
    return b;
}

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

    ll N, T, k, l;
    cin >> N >> T >> k >> l;
    mint iv = mint(6).inv(), a = iv*(k-1), b = iv*(l-k), c = iv*(7-l);
    vector<mint> dp(T+1);
    dp[1] = 1;
    for (int i=1; i<=T-1; i++){
        dp[i+1] += dp[i] * a;
        if (i != T-1) dp[i+2] += dp[i] * b;
    }

    if (N <= T){
        cout << dp[N].val() << endl;
        return 0;
    }

    /*
       M^(k)[0] * dpがT+kの答え
    */
    N -= T;
    vector M(T, vector<mint>(T));
    M[0][0] = a;
    M[0][1] = b;
    M[0][T-1] = c;
    for (int i=1; i<T; i++) M[i][i-1] = 1;
    M = pow(M, N);
    mint ans=0;
    for (int i=0; i<T; i++) ans += M[0][i] * dp[T-i];
    cout << ans.val() << endl;

    return 0;
}