Problem: 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.

Examples

Input

2 2


Output

3


Input

3 2


Output

19


Input

10 10


Output

211428932


Input

100000 50000


Output

3463133

Solution (in CPP):

Solution:
// C - GP 2
#include <bits/stdc++.h>

using namespace std;

#define	MOD 998244353

const int FACT_MAX = 2000000;
long long fact[FACT_MAX] = {1}, finv[FACT_MAX];

void init_fact()
{
	for (int i = 1; i < FACT_MAX; i++) fact[i] = fact[i - 1] * i % MOD;
	finv[FACT_MAX - 1] = 1;
	for (long long n = MOD - 2, a = fact[FACT_MAX - 1]; n > 0; a = a * a % MOD, n >>= 1)
		if (n & 1) finv[FACT_MAX - 1] = finv[FACT_MAX - 1] * a % MOD;
	for (int i = FACT_MAX - 2; i >= 0; i--)
		finv[i] = finv[i + 1] * (i + 1) % MOD;
}

long long cnr(int n, int r)
{
	if (r < 0 || r > n) return 0;
	if (r == 0 || r == n) return 1;
	return fact[n] * finv[r] % MOD * finv[n - r] % MOD;
}

int main(int argc, char *argv[])
{
	ios::sync_with_stdio(false); cin.tie(0);

	long long n, m;
	cin >> n >> m;

	init_fact();

	long long ans = 0;
	for (int k = 0; k <= min(m, n); k++) {
		if ((3 * m - k) % 2) continue;
		long long c = (3 * m - k) / 2;
		ans = (ans + cnr(c + n - 1, c) * cnr(n, k) % MOD) % MOD;
	}

	long long r = 0;
	for (int k = 0; k <= min(m, n); k++) {
		if ((m - k) % 2) continue;
		long long c = (m - k) / 2;
		r = (r + cnr(c + n - 1, c) * cnr(n, k) % MOD - cnr(c + n - 2, c) * cnr(n - 1, k) % MOD + MOD) % MOD;
	}
	ans = (ans - r * n % MOD + MOD) % MOD;

	cout << ans << endl;

	return 0;
}