#include <bits/stdc++.h>
using namespace std;

const int mod = 998244353;

#define N 3000030
int fac[N], rfac[N], ri[N];
void init() {
    fac[0] = 1;
    for (int i = 1; i < N; i ++) fac[i] = 1ll * fac[i-1] * i % mod;
    rfac[0] = rfac[1] = 1, ri[1] = 1;
    for(int i = 2; i < N; i ++) {
        int t = mod / i;
        ri[i] = (1ll * t * (mod - ri[mod % i])) % mod;
        rfac[i] = (1ll * rfac[i-1] * ri[i]) % mod;
    }
}

int C(int n, int k) {
	if (!(0 <= k && k <= n)) return 0;
	return 1ll * fac[n] * rfac[k] % mod * rfac[n-k] % mod;
}

int main() {
	init();
	int n, m;
	scanf("%d%d", &n, &m);
	int ans = 0;
	for (int o = m & 1; o <= m && o <= n; o += 2) {
		int d = (3 * m - o) / 2;
		ans = (ans + 1ll * C(n, o) * ((C(n + d - 1, d) - 1ll * o * C(n + d - m - 1, d - m) - 1ll * (n - o) * C(n+d-m-2,d-m-1)) % mod)) % mod;
	}
	if (ans < 0) ans += mod;
	cout<<ans<<endl;
}

### Prompt

Create a solution in Cpp for 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 mod = 998244353;

#define N 3000030
int fac[N], rfac[N], ri[N];
void init() {
    fac[0] = 1;
    for (int i = 1; i < N; i ++) fac[i] = 1ll * fac[i-1] * i % mod;
    rfac[0] = rfac[1] = 1, ri[1] = 1;
    for(int i = 2; i < N; i ++) {
        int t = mod / i;
        ri[i] = (1ll * t * (mod - ri[mod % i])) % mod;
        rfac[i] = (1ll * rfac[i-1] * ri[i]) % mod;
    }
}

int C(int n, int k) {
	if (!(0 <= k && k <= n)) return 0;
	return 1ll * fac[n] * rfac[k] % mod * rfac[n-k] % mod;
}

int main() {
	init();
	int n, m;
	scanf("%d%d", &n, &m);
	int ans = 0;
	for (int o = m & 1; o <= m && o <= n; o += 2) {
		int d = (3 * m - o) / 2;
		ans = (ans + 1ll * C(n, o) * ((C(n + d - 1, d) - 1ll * o * C(n + d - m - 1, d - m) - 1ll * (n - o) * C(n+d-m-2,d-m-1)) % mod)) % mod;
	}
	if (ans < 0) ans += mod;
	cout<<ans<<endl;
}

```