#include <bits/stdc++.h>

const int N = 3e6 + 5, djq = 998244353;

int n, m, fac[N], inv[N], ans;

int C(int n, int m)
{
	return 1ll * fac[n] * inv[m] % djq * inv[n - m] % djq;
}

int main()
{
	std::cin >> n >> m;
	fac[0] = inv[0] = inv[1] = 1;
	for (int i = 1; i < N; i++) fac[i] = 1ll * fac[i - 1] * i % djq;
	for (int i = 2; i < N; i++) inv[i] = 1ll * (djq - djq / i) * inv[djq % i] % djq;
	for (int i = 2; i < N; i++) inv[i] = 1ll * inv[i] * inv[i - 1] % djq;
	for (int i = 0; i <= m && i <= n; i++) if (!(m - i & 1))
		ans = (1ll * C(n, i) * C((m * 3 - i >> 1) + n - 1, n - 1) + ans) % djq;
	ans = (ans - 1ll * n * C(m + n - 2, n - 1) % djq + djq) % djq;
	return std::cout << ans << std::endl, 0;
}
### Prompt

Your challenge is to write a Cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>

const int N = 3e6 + 5, djq = 998244353;

int n, m, fac[N], inv[N], ans;

int C(int n, int m)
{
	return 1ll * fac[n] * inv[m] % djq * inv[n - m] % djq;
}

int main()
{
	std::cin >> n >> m;
	fac[0] = inv[0] = inv[1] = 1;
	for (int i = 1; i < N; i++) fac[i] = 1ll * fac[i - 1] * i % djq;
	for (int i = 2; i < N; i++) inv[i] = 1ll * (djq - djq / i) * inv[djq % i] % djq;
	for (int i = 2; i < N; i++) inv[i] = 1ll * inv[i] * inv[i - 1] % djq;
	for (int i = 0; i <= m && i <= n; i++) if (!(m - i & 1))
		ans = (1ll * C(n, i) * C((m * 3 - i >> 1) + n - 1, n - 1) + ans) % djq;
	ans = (ans - 1ll * n * C(m + n - 2, n - 1) % djq + djq) % djq;
	return std::cout << ans << std::endl, 0;
}
```