#include <bits/stdc++.h>

using namespace std;

const int MX = 3000000, md = 998244353;

long long inv[MX], f[MX], inv_f[MX];

long long c(int n, int k) {
	if (k == -1) return n == -1 ? 1 : 0;
	
	return f[n] * inv_f[n - k] % md * inv_f[k] % md;
}

int main() {
	inv[1] = f[0] = f[1] = inv_f[0] = inv_f[1] = 1;
	for (int i = 2; i < MX; i++) {
		inv[i] = md - inv[md % i] * (md / i) % md;
		f[i] = f[i - 1] * i % md;
		inv_f[i] = inv_f[i - 1] * inv[i] % md;
	}
	
	int n, m;
	ignore = scanf("%d %d", &n, &m);
	
	int ans = 0;
	for (int k = m % 2; k <= m && k <= n; k += 2) {
		ans = (ans + c(n, k) * c((3 * m - k) / 2 + n - 1, n - 1)) % md;
	}
	
	for (int k = 2 * m + 1; k <= 3 * m; k++) {
		ans = (ans - c(3 * m - k + n - 2, n - 2) * n) % md;
	}
	
	if (ans < 0) ans += md;
	
	printf("%d\n", ans);
	
	return 0;
}

### Prompt

Generate 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>

using namespace std;

const int MX = 3000000, md = 998244353;

long long inv[MX], f[MX], inv_f[MX];

long long c(int n, int k) {
	if (k == -1) return n == -1 ? 1 : 0;
	
	return f[n] * inv_f[n - k] % md * inv_f[k] % md;
}

int main() {
	inv[1] = f[0] = f[1] = inv_f[0] = inv_f[1] = 1;
	for (int i = 2; i < MX; i++) {
		inv[i] = md - inv[md % i] * (md / i) % md;
		f[i] = f[i - 1] * i % md;
		inv_f[i] = inv_f[i - 1] * inv[i] % md;
	}
	
	int n, m;
	ignore = scanf("%d %d", &n, &m);
	
	int ans = 0;
	for (int k = m % 2; k <= m && k <= n; k += 2) {
		ans = (ans + c(n, k) * c((3 * m - k) / 2 + n - 1, n - 1)) % md;
	}
	
	for (int k = 2 * m + 1; k <= 3 * m; k++) {
		ans = (ans - c(3 * m - k + n - 2, n - 2) * n) % md;
	}
	
	if (ans < 0) ans += md;
	
	printf("%d\n", ans);
	
	return 0;
}

```