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;
const double pi = acos(0.0) * 2.0;
const long double eps = 1e-10;
const int step[8][2] = {{-1, 0}, {0, 1}, {1, 0},  {0, -1},
                        {-1, 1}, {1, 1}, {1, -1}, {-1, -1}};
template <class T>
inline T abs1(T a) {
  return a < 0 ? -a : a;
}
template <typename t, typename t1>
t min1(t a, t1 b) {
  return a < b ? a : b;
}
template <typename t, typename... arg>
t min1(t a, arg... arr) {
  return min1(a, min1(arr...));
}
template <typename t, typename t1>
t max1(t a, t1 b) {
  return a > b ? a : b;
}
template <typename t, typename... arg>
t max1(t a, arg... arr) {
  return max1(a, max1(arr...));
}
inline int jud(double a, double b) {
  if (abs(a) < eps && abs(b) < eps)
    return 0;
  else if (abs1(a - b) / max(abs1(a), abs1(b)) < eps)
    return 0;
  if (a < b) return -1;
  return 1;
}
template <typename t>
inline int jud(t a, t b) {
  if (a < b) return -1;
  if (a == b) return 0;
  return 1;
}
template <typename it, typename t1>
inline int RangeFind(t1 val, it a, int na, bool f_small = 1, bool f_lb = 1) {
  if (na == 0) return 0;
  int be = 0, en = na - 1;
  if (*a <= *(a + na - 1)) {
    if (f_lb == 0)
      while (be < en) {
        int mid = (be + en + 1) / 2;
        if (jud(*(a + mid), val) != 1)
          be = mid;
        else
          en = mid - 1;
      }
    else
      while (be < en) {
        int mid = (be + en) / 2;
        if (jud(*(a + mid), val) != -1)
          en = mid;
        else
          be = mid + 1;
      }
    if (f_small && jud(*(a + be), val) == 1) be--;
    if (!f_small && jud(*(a + be), val) == -1) be++;
  } else {
    if (f_lb)
      while (be < en) {
        int mid = (be + en + 1) / 2;
        if (jud(*(a + mid), val) != -1)
          be = mid;
        else
          en = mid - 1;
      }
    else
      while (be < en) {
        int mid = (be + en) / 2;
        if (jud(*(a + mid), val) != 1)
          en = mid;
        else
          be = mid + 1;
      }
    if (!f_small && jud(*(a + be), val) == -1) be--;
    if (f_small && jud(*(a + be), val) == 1) be++;
  }
  return be;
}
template <class T>
inline T lowb(T num) {
  return num & (-num);
}
inline int bitnum(unsigned int nValue) { return __builtin_popcount(nValue); }
inline int bitnum(int nValue) { return __builtin_popcount(nValue); }
inline int bitnum(unsigned long long nValue) {
  return __builtin_popcountll(nValue);
}
inline int bitnum(long long nValue) { return __builtin_popcountll(nValue); }
inline int bitmaxl(unsigned int a) {
  if (a == 0) return 0;
  return 32 - __builtin_clz(a);
}
inline int bitmaxl(int a) {
  if (a == 0) return 0;
  return 32 - __builtin_clz(a);
}
inline int bitmaxl(unsigned long long a) {
  if (a == 0) return 0;
  return 64 - __builtin_clzll(a);
}
inline int bitmaxl(long long a) {
  if (a == 0) return 0;
  return 64 - __builtin_clzll(a);
}
long long pow(long long n, long long m, long long mod = 0) {
  if (m < 0) return 0;
  long long ans = 1;
  long long k = n;
  while (m) {
    if (m & 1) {
      ans *= k;
      if (mod) ans %= mod;
    }
    k *= k;
    if (mod) k %= mod;
    m >>= 1;
  }
  return ans;
}
template <class t1, class t2>
inline void add(t1 &a, t2 b, int mod = -1) {
  if (mod == -1) mod = 1000000007;
  a += b;
  while (a >= mod) a -= mod;
  while (a < 0) a += mod;
}
template <class t>
void output1(t arr) {
  for (int i = 0; i < (int)arr.size(); i++) cerr << arr[i] << ' ';
  cerr << endl;
}
template <class t>
void output2(t arr) {
  for (int i = 0; i < (int)arr.size(); i++) output1(arr[i]);
}
const int mod = 998244353;
const int maxn = 4001000;
long long fac[maxn], invfac[maxn], inv[maxn];
int n, m;
long long c(int a, int b) {
  if (b > a) return 0;
  return fac[a] * invfac[b] % mod * invfac[a - b] % mod;
}
long long GetB(int n, int a) {
  long long ra = c(n - a + m - 1, m - 1) * c(n + m, a) % mod * fac[a] % mod *
                 fac[n - a] % mod;
  long long rb = c(n + m, m) * fac[n] % mod;
  return rb * pow(rb + mod - ra, mod - 2, mod) % mod;
}
long long GetAll() {
  long long ans = 0, multi = 1;
  for (int i = 0; i <= n; i++) {
    ans += m * inv[n + m - i] % mod * multi % mod * (i + 1);
    ans %= mod;
    multi = multi * (n - i) % mod * inv[n + m - i] % mod;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  scanf("%d%d", &n, &m);
  fac[0] = 1;
  for (int i = 1; i < maxn; i++) {
    fac[i] = fac[i - 1] * i % mod;
  }
  invfac[maxn - 1] = pow(fac[maxn - 1], mod - 2, mod);
  for (int j = maxn - 2; j >= 0; j--) {
    invfac[j] = invfac[j + 1] * (j + 1) % mod;
    inv[j + 1] = fac[j] * invfac[j + 1] % mod;
  }
  long long ans = 0;
  for (int i = n - 1; i >= 0; i--) {
    long long flag = 1;
    if ((i - (n - 1)) % 2) flag = mod - 1;
    ans += flag * c(n, i) % mod * GetB(n, i);
    ans %= mod;
  }
  cout << ans * GetAll() % mod << endl;
  return 0;
}