Problem: zscoder has a deck of n+m custom-made cards, which consists of n cards labelled from 1 to n and m jokers. Since zscoder is lonely, he wants to play a game with himself using those cards. 

Initially, the deck is shuffled uniformly randomly and placed on the table. zscoder has a set S which is initially empty. 

Every second, zscoder draws the top card from the deck. 

  * If the card has a number x written on it, zscoder removes the card and adds x to the set S. 
  * If the card drawn is a joker, zscoder places all the cards back into the deck and reshuffles (uniformly randomly) the n+m cards to form a new deck (hence the new deck now contains all cards from 1 to n and the m jokers). Then, if S currently contains all the elements from 1 to n, the game ends. Shuffling the deck doesn't take time at all. 



What is the expected number of seconds before the game ends? We can show that the answer can be written in the form P/Q where P, Q are relatively prime integers and Q ≠ 0 mod 998244353. Output the value of (P ⋅ Q^{-1}) modulo 998244353.

Input

The only line of the input contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^{6}).

Output

Output a single integer, the value of (P ⋅ Q^{-1}) modulo 998244353.

Examples

Input


2 1


Output


5


Input


3 2


Output


332748127


Input


14 9


Output


969862773

Note

For the first sample, it can be proven that the expected time before the game ends is 5 seconds.

For the second sample, it can be proven that the expected time before the game ends is 28/3 seconds.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using uint = unsigned int;
template <class T>
using V = vector<T>;
template <class T>
using VV = vector<vector<T>>;
template <class T, class U>
void chmax(T& x, U y) {
  if (x < y) x = y;
}
template <class T, class U>
void chmin(T& x, U y) {
  if (y < x) x = y;
}
template <class T>
void mkuni(V<T>& v) {
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
}
template <class S, class T>
ostream& operator<<(ostream& o, const pair<S, T>& p) {
  return o << "(" << p.first << "," << p.second << ")";
}
template <class T>
ostream& operator<<(ostream& o, const vector<T>& vc) {
  o << "{";
  for (const T& v : vc) o << v << ",";
  o << "}";
  return o;
}
constexpr ll TEN(int n) { return (n == 0) ? 1 : 10 * TEN(n - 1); }
template <unsigned int mod_>
struct ModInt {
  using uint = unsigned int;
  using ll = long long;
  using ull = unsigned long long;
  constexpr static uint mod = mod_;
  uint v;
  ModInt() : v(0) {}
  ModInt(ll _v) : v(normS(_v % mod + mod)) {}
  explicit operator bool() const { return v != 0; }
  static uint normS(const uint& x) { return (x < mod) ? x : x - mod; }
  static ModInt make(const uint& x) {
    ModInt m;
    m.v = x;
    return m;
  }
  ModInt operator+(const ModInt& b) const { return make(normS(v + b.v)); }
  ModInt operator-(const ModInt& b) const { return make(normS(v + mod - b.v)); }
  ModInt operator-() const { return make(normS(mod - v)); }
  ModInt operator*(const ModInt& b) const { return make((ull)v * b.v % mod); }
  ModInt operator/(const ModInt& b) const { return *this * b.inv(); }
  ModInt& operator+=(const ModInt& b) { return *this = *this + b; }
  ModInt& operator-=(const ModInt& b) { return *this = *this - b; }
  ModInt& operator*=(const ModInt& b) { return *this = *this * b; }
  ModInt& operator/=(const ModInt& b) { return *this = *this / b; }
  ModInt& operator++(int) { return *this = *this + 1; }
  ModInt& operator--(int) { return *this = *this - 1; }
  ll extgcd(ll a, ll b, ll& x, ll& y) const {
    ll p[] = {a, 1, 0}, q[] = {b, 0, 1};
    while (*q) {
      ll t = *p / *q;
      for (int i = 0; i < int(3); i++) swap(p[i] -= t * q[i], q[i]);
    }
    if (p[0] < 0)
      for (int i = 0; i < int(3); i++) p[i] = -p[i];
    x = p[1], y = p[2];
    return p[0];
  }
  ModInt inv() const {
    ll x, y;
    extgcd(v, mod, x, y);
    return make(normS(x + mod));
  }
  ModInt pow(ll p) const {
    if (p < 0) return inv().pow(-p);
    ModInt a = 1;
    ModInt x = *this;
    while (p) {
      if (p & 1) a *= x;
      x *= x;
      p >>= 1;
    }
    return a;
  }
  bool operator==(const ModInt& b) const { return v == b.v; }
  bool operator!=(const ModInt& b) const { return v != b.v; }
  friend istream& operator>>(istream& o, ModInt& x) {
    ll tmp;
    o >> tmp;
    x = ModInt(tmp);
    return o;
  }
  friend ostream& operator<<(ostream& o, const ModInt& x) { return o << x.v; }
};
using mint = ModInt<998244353>;
V<mint> fact, ifact;
mint Choose(int a, int b) {
  if (b < 0 || a < b) return 0;
  return fact[a] * ifact[b] * ifact[a - b];
}
void InitFact(int N) {
  fact.resize(N);
  ifact.resize(N);
  fact[0] = 1;
  for (int i = 1; i <= int(N - 1); i++) fact[i] = fact[i - 1] * i;
  ifact[N - 1] = fact[N - 1].inv();
  for (int i = N - 2; i >= 0; i--) ifact[i] = ifact[i + 1] * (i + 1);
}
void test() {
  int N, M;
  cin >> N >> M;
  V<mint> f(N + 1);
  for (int x = int(N) - 1; x >= 0; x--) {
    void(0);
    mint sum_pi = 0;
    V<mint> p(N - x + 1), q(N - x + 1);
    for (int i = 0; i < int(N - x + 1); i++) {
      mint pi = 0;
      mint qi = 0;
      for (int j = 0; j < int(x + 1); j++) {
        mint pij = ifact[N + M] * fact[N + M - (i + j) - 1] * M * fact[i + j] *
                   Choose(N - x, i) * Choose(x, j);
        pi += pij;
        qi += pij * (i + j + 1);
      }
      p[i] = pi, q[i] = qi;
      void(0);
      void(0);
      void(0);
      sum_pi += pi;
    }
    {
      mint sum = 0;
      for (int i = 0; i < int(N - x + 1); i++) {
        sum += q[i];
        if (i) sum += p[i] * f[x + i];
      }
      f[x] = sum / (mint(1) - p[0]);
    }
    assert(sum_pi == 1);
  }
  V<mint> df(N);
  for (int i = 0; i < int(N); i++) df[i] = f[i] - f[i + 1];
  reverse(df.begin(), df.end());
  for (int i = int(N) - 1; i >= 0; i--) df[i] /= df[0];
  cout << df << endl;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(20);
  InitFact(3000000);
  int N, M;
  cin >> N >> M;
  mint ans = 0;
  for (int i = 0; i < int(N); i++) {
    if (i == 0)
      ans += mint(M).inv();
    else
      ans += (mint(M + 1) * (i + 1)).inv();
  }
  ans *= M;
  ans *= N + M + 1;
  cout << ans << endl;
}