#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
const int Mod=998244353;
int n,m;
#define Maxn 3000010
int fact[Maxn],inv[Maxn];
inline int C(int i,int j){return 1ll*fact[i]*inv[i-j]%Mod*inv[j]%Mod;}
inline int calc(int x){return C(x+n-1,n-1);}
int main(){
	scanf("%d%d",&n,&m);
	fact[0]=1;
	for(register int i=1;i<=3000000;++i)fact[i]=1ll*fact[i-1]*i%Mod;
	inv[0]=inv[1]=1;
	for(register int i=2;i<=3000000;++i)
		inv[i]=1ll*(Mod-Mod/i)*inv[Mod%i]%Mod;
	for(register int i=2;i<=3000000;++i)inv[i]=1ll*inv[i-1]*inv[i]%Mod;
	int Ans=0;
	for(register int i=0;i<=min(m,n);++i)
	if((m-i)%2==0){
		int at=(3*m-i)/2;
		int res=calc(at);
		res=(res-1ll*i*C(at-m+n-1,n-1)%Mod+Mod)%Mod;
		if(at>m)res=(res-1ll*(n-i)*C(at-m+n-2,n-1)%Mod+Mod)%Mod;
		res=1ll*res*C(n,i)%Mod;
		Ans=(Ans+res)%Mod;
	}
	printf("%d\n",Ans);
	return 0;
}
### Prompt

Please formulate 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<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
const int Mod=998244353;
int n,m;
#define Maxn 3000010
int fact[Maxn],inv[Maxn];
inline int C(int i,int j){return 1ll*fact[i]*inv[i-j]%Mod*inv[j]%Mod;}
inline int calc(int x){return C(x+n-1,n-1);}
int main(){
	scanf("%d%d",&n,&m);
	fact[0]=1;
	for(register int i=1;i<=3000000;++i)fact[i]=1ll*fact[i-1]*i%Mod;
	inv[0]=inv[1]=1;
	for(register int i=2;i<=3000000;++i)
		inv[i]=1ll*(Mod-Mod/i)*inv[Mod%i]%Mod;
	for(register int i=2;i<=3000000;++i)inv[i]=1ll*inv[i-1]*inv[i]%Mod;
	int Ans=0;
	for(register int i=0;i<=min(m,n);++i)
	if((m-i)%2==0){
		int at=(3*m-i)/2;
		int res=calc(at);
		res=(res-1ll*i*C(at-m+n-1,n-1)%Mod+Mod)%Mod;
		if(at>m)res=(res-1ll*(n-i)*C(at-m+n-2,n-1)%Mod+Mod)%Mod;
		res=1ll*res*C(n,i)%Mod;
		Ans=(Ans+res)%Mod;
	}
	printf("%d\n",Ans);
	return 0;
}
```