#include<cstdio>
#define RI register int
#define CI const int&
using namespace std;
const int N=3000005,mod=998244353;
int n,m,fact[N],inv[N],ans,tp;
inline void inc(int& x,CI y)
{
	if ((x+=y)>=mod) x-=mod;
}
inline void dec(int& x,CI y)
{
	if ((x-=y)<0) x+=mod;
}
inline int quick_pow(int x,int p=mod-2,int mul=1)
{
	for (;p;p>>=1,x=1LL*x*x%mod) if (p&1) mul=1LL*mul*x%mod; return mul;
}
inline int C(CI n,CI m)
{
	if (n<m) return 0; return 1LL*fact[n]*inv[m]%mod*inv[n-m]%mod;
}
inline void init(CI n)
{
	RI i; for (fact[0]=i=1;i<=n;++i) fact[i]=1LL*fact[i-1]*i%mod;
	for (inv[n]=quick_pow(fact[n]),i=n-1;~i;--i) inv[i]=1LL*inv[i+1]*(i+1)%mod;
}
int main()
{
	RI i; scanf("%d%d",&n,&m); init(n+3*m);
	for (i=0;i<=m;++i) if (!((3*m-i)&1))
	tp=(3*m-i)/2,inc(ans,1LL*C(n,i)*C(tp+n-1,n-1)%mod);
	for (i=2*m+1;i<=3*m;++i) dec(ans,1LL*n*C(3*m-i+n-2,n-2)%mod);
	return printf("%d",ans),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<cstdio>
#define RI register int
#define CI const int&
using namespace std;
const int N=3000005,mod=998244353;
int n,m,fact[N],inv[N],ans,tp;
inline void inc(int& x,CI y)
{
	if ((x+=y)>=mod) x-=mod;
}
inline void dec(int& x,CI y)
{
	if ((x-=y)<0) x+=mod;
}
inline int quick_pow(int x,int p=mod-2,int mul=1)
{
	for (;p;p>>=1,x=1LL*x*x%mod) if (p&1) mul=1LL*mul*x%mod; return mul;
}
inline int C(CI n,CI m)
{
	if (n<m) return 0; return 1LL*fact[n]*inv[m]%mod*inv[n-m]%mod;
}
inline void init(CI n)
{
	RI i; for (fact[0]=i=1;i<=n;++i) fact[i]=1LL*fact[i-1]*i%mod;
	for (inv[n]=quick_pow(fact[n]),i=n-1;~i;--i) inv[i]=1LL*inv[i+1]*(i+1)%mod;
}
int main()
{
	RI i; scanf("%d%d",&n,&m); init(n+3*m);
	for (i=0;i<=m;++i) if (!((3*m-i)&1))
	tp=(3*m-i)/2,inc(ans,1LL*C(n,i)*C(tp+n-1,n-1)%mod);
	for (i=2*m+1;i<=3*m;++i) dec(ans,1LL*n*C(3*m-i+n-2,n-2)%mod);
	return printf("%d",ans),0;
}
```