#include <atcoder/all>
#include <bits/stdc++.h>
#define rep(i, a, b) for (ll i = (ll)(a); i < (ll)(b); i++)
using namespace atcoder;
using namespace std;

typedef long long ll;

using mint = modint998244353;

struct Comb {
    vector<mint> fact, ifact;
    int MAX_COM;
    Comb() {}
    Comb(int n) {
        MAX_COM = n; // ここでMAX入力を調整
        init(998244353, MAX_COM);
    }
    void init(long long MOD, long long MAX_COM) {
        int n = MAX_COM;
        assert(n < MOD);
        fact = vector<mint>(n + 1);
        ifact = vector<mint>(n + 1);
        fact[0] = 1;
        for (int i = 1; i <= n; ++i)
            fact[i] = fact[i - 1] * i;
        ifact[n] = fact[n].inv();
        for (int i = n; i >= 1; --i)
            ifact[i - 1] = ifact[i] * i;
    }
    mint operator()(long long n, long long k) {
        if (k < 0 || k > n)
            return 0;
        return fact[n] * ifact[k] * ifact[n - k];
    }
};
Comb comb(5000010);

struct root5 {
    // a + b * sqrt(5)
    mint a, b;
    root5(mint a, mint b) : a(a), b(b) {}
    root5() {}
    root5 operator+(const root5 &r) const { return root5(a + r.a, b + r.b); }
    root5 operator-(const root5 &r) const { return root5(a - r.a, b - r.b); }
    root5 operator*(const root5 &r) const {
        return root5(a * r.a + mint(5) * b * r.b, a * r.b + b * r.a);
    }
    root5 operator*(const ll n) const { return root5(a * n, b * n); }
    root5 operator/(const root5 &r) const {
        root5 rr = root5(r.a, -r.b);
        root5 nr = r * rr;
        root5 res = *this * rr;
        res = res / nr.a.val();
        return res;
    }
    root5 operator/(const ll n) const {
        mint x = a / n;
        mint y = b / n;
        return root5(x, y);
    }
    root5 pow(ll n) {
        root5 x = *this;
        root5 res = root5(1, 0);
        while (n) {
            if (n & 1)
                res = res * x;
            x = x * x;
            n >>= 1;
        }
        return res;
    }
};

void solve(ll n, ll k) {
    root5 ans = root5(0, 0);
    root5 a = root5((mint)1 / 2, (mint)1 / 2);
    root5 b = root5((mint)1 / 2, (mint)-1 / 2);
    root5 one = root5(1, 0);
    rep(i, 0, k + 1) {
        root5 p(1, 0);
        p = p * comb(k, i).val();
        if (i % 2)
            p = p * -1;
        root5 x = a.pow(k - i) * b.pow(i);
        if (x.a.val() == 1 && x.b.val() == 0) {
            ans = ans + root5(n, 0) * p;
        } else {
            ans = ans + p * (x.pow(n + 1) - x) / (x - one);
        }
    }
    ans = ans / root5(0, (mint)1).pow(k);
    cout << ans.a.val() << endl;
    return;
}

int main() {
    // rep(i, 1, 10) { solve(i, 3); }
    ll n;
    int k;
    cin >> n >> k;
    solve(n, k);
}