description: There are n heroes fighting in the arena. Initially, the i-th hero has a_i health points.
  
  The fight in the arena takes place in several rounds. At the beginning of each round, each alive hero deals 1 damage to all other heroes. Hits of all heroes occur simultaneously. Heroes whose health is less than 1 at the end of the round are considered killed.
  
  If exactly 1 hero remains alive after a certain round, then he is declared the winner. Otherwise, there is no winner.
  
  Your task is to calculate the number of ways to choose the initial health points for each hero a_i, where 1 ≤ a_i ≤ x, so that there is no winner of the fight. The number of ways can be very large, so print it modulo 998244353. Two ways are considered different if at least one hero has a different amount of health. For example, [1, 2, 1] and [2, 1, 1] are different.
  
  Input
  
  The only line contains two integers n and x (2 ≤ n ≤ 500; 1 ≤ x ≤ 500).
  
  Output
  
  Print one integer — the number of ways to choose the initial health points for each hero a_i, where 1 ≤ a_i ≤ x, so that there is no winner of the fight, taken modulo 998244353. 
  
  Examples
  
  Input
  
  
  2 5
  
  
  Output
  
  
  5
  
  
  Input
  
  
  3 3
  
  
  Output
  
  
  15
  
  
  Input
  
  
  5 4
  
  
  Output
  
  
  1024
  
  
  Input
  
  
  13 37
  
  
  Output
  
  
  976890680
solution: #include <bits/stdc++.h>
  using namespace std;
  const int intmax = 2147483647;
  const int intmin = -2147483648;
  const int mod = 1000000007;
  const int mod2 = 998244353;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  template <typename T>
  istream &operator>>(istream &in, vector<T> &v) {
    for (auto &i : v) in >> i;
    return in;
  }
  template <typename T>
  ostream &operator<<(ostream &ost, vector<T> v) {
    if (&ost == &cerr) {
      cerr << "{";
      int cnt = v.size();
      for (auto x : v) {
        cerr << x;
        cnt--;
        if (cnt) cerr << ", ";
      }
      cerr << "}";
    } else
      for (auto i : v) ost << i << " ";
    return ost;
  }
  template <typename T>
  ostream &operator<<(ostream &ost, set<T> s) {
    if (&ost == &cerr) {
      cerr << "{";
      int cnt = s.size();
      for (auto x : s) {
        cerr << x;
        cnt--;
        if (cnt) cerr << ", ";
      }
      cerr << "}";
    } else
      for (auto i : s) ost << i << " ";
    return ost;
  }
  template <typename U, typename V>
  ostream &operator<<(ostream &ost, map<U, V> &m) {
    if (&ost == &cerr) {
      cerr << "{";
      int cnt = m.size();
      for (auto x : m) {
        cerr << x;
        cnt--;
        if (cnt) cerr << ", ";
      }
      cerr << "}";
    } else
      for (auto i : m) ost << i;
    return ost;
  }
  template <typename U, typename V>
  istream &operator>>(istream &in, pair<U, V> &p) {
    in >> p.first >> p.second;
    return in;
  }
  template <typename U, typename V>
  ostream &operator<<(ostream &ost, pair<U, V> p) {
    if (&ost == &cerr)
      ost << "{" << p.first << ": " << p.second << "}";
    else
      ost << p.first << " " << p.second << " ";
    return ost;
  }
  template <typename... T>
  void read(T &...args) {
    ((cin >> args), ...);
  }
  template <typename... T>
  void print(T... args) {
    ((cout << args << " "), ...);
    cout << '\n';
  }
  long long binExpp(long long a, long long b, long long modulo) {
    long long prod = a;
    long long ans = 1;
    while (b) {
      if (b % 2) ans = (ans * prod) % modulo;
      b /= 2;
      prod = (prod * prod) % modulo;
    }
    return ans;
  }
  template <const int mod>
  class Mint {
    long long val;
  
   public:
    Mint(long long _val) { val = (_val % mod + mod) % mod; }
    Mint() { val = 0; }
    Mint operator+(Mint b) { return Mint((this->val + b.val) % mod); }
    Mint operator-(Mint b) { return Mint(this->val - b.val + mod) % mod; }
    Mint operator*(Mint b) { return Mint((this->val * b.val) % mod); }
    Mint operator/(Mint b) {
      return (Mint(binExpp(b(), mod - 2, mod)) * (*this));
    }
    long long &operator()() { return val; }
    Mint operator+=(Mint b) { return *this = *this + b; }
    Mint operator-=(Mint b) { return *this = *this - b; }
    Mint operator*=(Mint b) { return *this = *this * b; }
    Mint operator/=(Mint b) { return *this = *this / b; }
    int operator%(int modulo) { return val % modulo; }
    int operator%(Mint modulo) { return val % (modulo()); }
    Mint operator++() { return *this = *this + Mint(1); }
    Mint operator=(int x) { return (*this).val = (x % mod + mod) % mod; }
    Mint operator^(Mint pow) { return Mint(binExpp((*this)(), pow(), mod)); }
  };
  template <const int mod>
  istream &operator>>(istream &in, Mint<mod> &a) {
    return in >> a();
  }
  template <const int mod>
  ostream &operator<<(ostream &outt, Mint<mod> a) {
    return outt << a();
  }
  template <const int mod>
  Mint<mod> operator+(int mul, Mint<mod> m) {
    return m + mul;
  }
  template <const int mod>
  Mint<mod> operator*(int mul, Mint<mod> m) {
    return m * Mint<mod>(mul);
  }
  template <const int mod>
  vector<Mint<mod>> Factorial(int n) {
    vector<Mint<mod>> fact(n + 1);
    fact[0] = 1;
    for (int i = 1; i <= n; i++) fact[i] = i * fact[i - 1];
    return fact;
  }
  using mint = Mint<1000000007>;
  using mint2 = Mint<998244353>;
  using vmi = vector<mint>;
  auto fact = Factorial<mod2>(100000);
  inline mint2 C(mint2 n, mint2 r) {
    return fact[n()] / (fact[r()] * fact[(n - r)()]);
  }
  mint2 rec(mint2 n, mint2 x, vector<vector<int>> &dp) {
    if (dp[n()][x()] != -1) return dp[n()][x()];
    mint2 res = 0;
    res += x() / (n() - 1) * ((n - 1) ^ n);
    if (x() % (n() - 1)) res += mint2(x() % (n() - 1)) ^ n;
    if (x() > n() - 1) {
      for (int j = 2; j < n(); j++) {
        mint2 ways = 1;
        mint2 f = 0;
        for (int k = 1; k < x() / (n() - 1) + 1; k++) {
          f = ((n - 1) ^ (n - j));
          ways = C(n, j) * (f);
          res += ways * rec(j, x() - k * (n() - 1), dp);
        }
      }
    }
    return dp[n()][x()] = res();
  }
  void solve() {
    int n, x;
    cin >> n >> x;
    vector<vector<int>> dp(n + 1, vector<int>(x + 1, -1));
    cout << rec(n, x, dp) << '\n';
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t = 1;
    for (int var = 1; var < t + 1; var++) solve();
    return 0;
  }
  