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

using namespace std;
using namespace atcoder;

struct Fast {
  Fast() {
    std::cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << setprecision(10);
  }
} fast;

#define all(a) (a).begin(), (a).end()
#define contains(a, x) ((a).find(x) != (a).end())
#define rep(i, a, b) for (int i = (a); i < (int)(b); i++)
#define rrep(i, a, b) for (int i = (int)(b)-1; i >= (a); i--)
#define writejoin(s, a) rep(_i, 0, (a).size()) cout << (a)[_i] << (_i + 1 < (int)(a).size() ? s : "\n");
#define YN(b) cout << ((b) ? "YES" : "NO") << "\n";
#define Yn(b) cout << ((b) ? "Yes" : "No") << "\n";
#define yn(b) cout << ((b) ? "yes" : "no") << "\n";

using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vl = vector<ll>;
using mint = modint998244353;
using vm = vector<mint>;

template <typename mint>
class Factorial {
 public:
  Factorial(int max) : n(max) {
    f = std::vector<mint>(n + 1);
    finv = std::vector<mint>(n + 1);
    f[0] = 1;
    for (int i = 1; i <= n; i++) f[i] = f[i - 1] * i;
    finv[n] = f[n].inv();
    for (int i = n; i > 0; i--) finv[i - 1] = finv[i] * i;
  }
  mint fact(int k) const {
    assert(0 <= k && k <= n);
    return f[k];
  }
  mint fact_inv(int k) const {
    assert(0 <= k && k <= n);
    return finv[k];
  }
  mint binom(int k, int r) const {
    assert(0 <= k && k <= n);
    if (r < 0 || r > k) return 0;
    return f[k] * finv[r] * finv[k - r];
  }
  mint inv(int k) const {
    assert(0 < k && k <= n);
    return finv[k] * f[k - 1];
  }

 private:
  int n;
  std::vector<mint> f, finv;
};

using ar = array<mint, 3>;

ar mul(ar a, ar b) {
  ar r{0, 0, 0};
  rep(i, 0, 3) rep(j, 0, 3) r[(i + j) % 3] += a[i] * b[j];
  return r;
}

ar pow(ar a, int n) {
  ar r{1, 0, 0};
  while (n) {
    if (n & 1) r = mul(r, a);
    a = mul(a, a);
    n /= 2;
  }
  return r;
}

void solve() {
  int n, m;
  cin >> n >> m;

  Factorial<mint> fact(m);
  mint ans = 0;
  rep(a, 0, m + 1) rep(b, 0, m + 1 - a) {
    int c = m - a - b;
    ar v{0, 0, 0};
    int d = (2 * b + c) % 3;
    v[d] = fact.fact_inv(a) * fact.fact_inv(b) * fact.fact_inv(c);
    v = mul(v, pow(ar{a, b, c}, n));
    ans += v[0] - v[2];
  }
  ans *= fact.fact(m);
  ans /= mint(3).pow(m);
  cout << ans.val() << endl;
}

int main() {
  int t = 1;
  // cin >> t;
  while (t--) solve();
}