#include <cstdio>

const int p=998244353;
const int N=3000000;

int fac[5000000],inv[5000000];
int i,m,n,s;

inline int power(int x,int y)
{
	int s=1;
	for (int i=30;i>=0;i--)
	{
		s=1LL*s*s%p;
		if ((1<<i)&y)
			s=1LL*s*x%p;
	}
	return s;
}

inline void pre_C()
{
	fac[0]=1;
	for (int i=1;i<=N;i++)
		fac[i]=1LL*fac[i-1]*i%p;
	inv[N]=power(fac[N],p-2);
	for (int i=N-1;i>=0;i--)
		inv[i]=1LL*inv[i+1]*(i+1)%p;
	return;
}

inline int C(int x,int y)
{
	if ((y<0) || (y>x))
		return 0;
	else
		return 1LL*fac[x]*inv[y]%p*inv[x-y]%p;
}

int main()
{
	pre_C();
	scanf("%d%d",&n,&m);
	for (i=m;i<=m+m/2;i++)
		s=(s+1LL*C(i+n-1,n-1)*C(n,3*m-2*i))%p;
	for (i=m+1;i<=m+m/2;i++)
		s=(s-1LL*C(i-m+n-2,n-1)*C(n,3*m-2*i)%p*n%p+p)%p;
	for (i=m;i<=m+m/2;i++)
		s=(s-1LL*C(i-m+n-2,n-2)*C(n-1,3*m-2*i-1)%p*n%p+p)%p;
	printf("%d",s);
	return 0;
}
### Prompt

Your task is to create 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 <cstdio>

const int p=998244353;
const int N=3000000;

int fac[5000000],inv[5000000];
int i,m,n,s;

inline int power(int x,int y)
{
	int s=1;
	for (int i=30;i>=0;i--)
	{
		s=1LL*s*s%p;
		if ((1<<i)&y)
			s=1LL*s*x%p;
	}
	return s;
}

inline void pre_C()
{
	fac[0]=1;
	for (int i=1;i<=N;i++)
		fac[i]=1LL*fac[i-1]*i%p;
	inv[N]=power(fac[N],p-2);
	for (int i=N-1;i>=0;i--)
		inv[i]=1LL*inv[i+1]*(i+1)%p;
	return;
}

inline int C(int x,int y)
{
	if ((y<0) || (y>x))
		return 0;
	else
		return 1LL*fac[x]*inv[y]%p*inv[x-y]%p;
}

int main()
{
	pre_C();
	scanf("%d%d",&n,&m);
	for (i=m;i<=m+m/2;i++)
		s=(s+1LL*C(i+n-1,n-1)*C(n,3*m-2*i))%p;
	for (i=m+1;i<=m+m/2;i++)
		s=(s-1LL*C(i-m+n-2,n-1)*C(n,3*m-2*i)%p*n%p+p)%p;
	for (i=m;i<=m+m/2;i++)
		s=(s-1LL*C(i-m+n-2,n-2)*C(n-1,3*m-2*i-1)%p*n%p+p)%p;
	printf("%d",s);
	return 0;
}
```