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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

import os
import sys

if os.getenv("LOCAL"):
    sys.stdin = open("_in.txt", "r")

sys.setrecursionlimit(2147483647)
INF = float("inf")
IINF = 10 ** 18
MOD = 998244353

N, M = list(map(int, sys.stdin.readline().split()))


def get_factorials(max, mod=None):
    """
    階乗 0!, 1!, 2!, ..., max!
    :param int max:
    :param int mod:
    :return:
    """
    ret = [1]
    n = 1
    if mod:
        for i in range(1, max + 1):
            n *= i
            n %= mod
            ret.append(n)
    else:
        for i in range(1, max + 1):
            n *= i
            ret.append(n)
    return ret


factorials = get_factorials(M * 3 + N - 1, MOD)


# invs = mod_invs(M * 3 + N - 1, MOD)
# fi = [0, invs[1]]
# a = invs[1]
# for i in invs[2:]:
#     a = a * i % MOD
#     fi.append(a)
#
# print(invs)


def mod_inv(a, mod):
    """
    a の逆元
    :param int a:
    :param int mod:
    :return:
    """
    return pow(a, mod - 2, mod)


def ncr(n, r, mod=None):
    """
    scipy.misc.comb または scipy.special.comb と同じ
    組み合わせの数 nCr
    :param int n:
    :param int r:
    :param int mod: 3 以上の素数であること
    :rtype: int
    """
    if n < r:
        return 0

    return factorials[n] * mod_inv(factorials[r], mod) * mod_inv(factorials[n - r], mod) % mod


# 合計が M*3 となる数列の数
c1 = ncr(M * 3 + N - 1, N - 1, MOD)
# max が M*2 より大きい数列の数を引く
c2 = 0
for m in range(M * 2 + 1, M * 3 + 1):
    s = M * 3 - m
    c2 += ncr(s + N - 2, N - 2, MOD) * N
    c2 %= MOD
print((c1 - c2) % MOD)