Problem:
Score : 900 points

Problem Statement

We have a sequence of N integers: x=(x_0,x_1,\cdots,x_{N-1}).
Initially, x_i=0 for each i (0 \leq i \leq N-1).

Snuke will perform the following operation exactly M times:

Choose two distinct indices i, j (0 \leq i,j \leq N-1,\ i \neq j).
Then, replace x_i with x_i+2 and x_j with x_j+1.

Find the number of different sequences that can result after M operations.
Since it can be enormous, compute the count modulo 998244353.

Constraints

2 \leq N \leq 10^6

1 \leq M \leq 5 \times 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M

Output

Print the number of different sequences that can result after M operations, modulo 998244353.

Sample Input 1

2 2

Sample Output 1

3

After two operations, there are three possible outcomes:

x=(2,4)

x=(3,3)

x=(4,2)

For example, x=(3,3) can result after the following sequence of operations:

First, choose i=0,j=1, changing x from (0,0) to (2,1).

Second, choose i=1,j=0, changing x from (2,1) to (3,3).

Sample Input 2

3 2

Sample Output 2

19

Sample Input 3

10 10

Sample Output 3

211428932

Sample Input 4

100000 50000

Sample Output 4

3463133

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
using lint = long long;
template<class T = int> using V = vector<T>;
template<class T = int> using VV = V< V<T> >;

template<unsigned P> struct ModInt {
  using M = ModInt;
  unsigned v;
  ModInt() : v(0) {}
  ModInt(auto x) : v(x >= 0 ? x % P : (P - -x % P) % P) {}
  constexpr ModInt(unsigned v, int) : v(v) {}
  static constexpr unsigned p() { return P; }
  M operator+() const { return *this; }
  M operator-() const { return {v ? P - v : 0, 0}; }
  explicit operator bool() const noexcept { return v; }
  bool operator!() const noexcept { return !(bool)*this; }
  M operator*(M r) const { return M(*this) *= r; }
  M operator/(M r) const { return M(*this) /= r; }
  M operator+(M r) const { return M(*this) += r; }
  M operator-(M r) const { return M(*this) -= r; }
  bool operator==(M r) const { return v == r.v; }
  bool operator!=(M r) const { return !(*this == r); }
  M& operator*=(M r) { v = (uint64_t)v * r.v % P; return *this; }
  M& operator/=(M r) { return *this *= r.inv(); }
  M& operator+=(M r) { if ((v += r.v) >= P) v -= P; return *this; }
  M& operator-=(M r) { if ((v += P - r.v) >= P) v -= P; return *this; }
  M inv() const {
    int a = v, b = P, x = 1, u = 0;
    while (b) {
      int q = a / b;
      swap(a -= q * b, b);
      swap(x -= q * u, u);
    }
    assert(a == 1);
    return x;
  }
  M pow(auto n) const {
    if (n < 0) return pow(-n).inv();
    M res = 1;
    for (M a = *this; n; a *= a, n >>= 1) if (n & 1) res *= a;
    return res;
  }
  friend M operator*(auto l, M r) { return M(l) *= r; }
  friend M operator/(auto l, M r) { return M(l) /= r; }
  friend M operator+(auto l, M r) { return M(l) += r; }
  friend M operator-(auto l, M r) { return M(l) -= r; }
  friend ostream& operator<<(ostream& os, M r) { return os << r.v; }
  friend istream& operator>>(istream& is, M& r) { lint x; is >> x; r = x; return is; }
  friend bool operator==(auto l, M r) { return M(l) == r; }
  friend bool operator!=(auto l, M r) { return !(l == r); }
};
using Mint = ModInt<998244353>;

V<Mint> fact, ifact, inv, powB;
void init(int n, int B = 2) {
  fact.resize(n + 1);
  fact[0] = 1;
  for (int i = 1; i <= n; ++i) {
    fact[i] = i * fact[i - 1];
  }
  ifact.resize(n + 1);
  ifact[n] = 1 / fact[n];
  for (int i = n; i > 0; --i) {
    ifact[i - 1] = i * ifact[i];
  }
  inv.resize(n + 1);
  inv[1] = 1;
  for (int i = 2; i <= n; ++i) {
    int q = Mint::p() / i;
    inv[i] = -q * inv[Mint::p() - i * q];
  }
  powB.resize(n + 1);
  powB[0] = 1;
  for (int i = 0; i < n; ++i) {
    powB[i + 1] = powB[i] * B;
  }
}
Mint comb(int n, int r) {
  if (r < 0 or r > n) return 0;
  return fact[n] * ifact[r] * ifact[n - r];
}
Mint fn(int s, int k) {
  return comb(s + k - 1, s);
}

int main() {
  cin.tie(nullptr); ios::sync_with_stdio(false);
  int n, m; cin >> n >> m;
  init(n + 3 * m);
  Mint res = fn(3 * m, n);
  for (int i = 2 * m + 1; i <= 3 * m; ++i) {
    res -= n * fn(3 * m - i, n - 1);
  }
  for (int i = m + 2; i <= min(n, 3 * m); i += 2) {
    res -= fn(3 * m - i >> 1, n) * comb(n, i);
  }
  cout << res << '\n';
}