#include<iostream>
#define int long long
using namespace std;
const int mod=998244353;
const int N=3e6+2;
int fact[N],inv[N];
int binpow(int x,int y){
	int tich=1;
	while(y){
		if((y&1)){
			tich*=x;
			tich%=mod;
		}
		x*=x;
		x%=mod;
		y>>=1;
	}
	return tich;
}
int C(int x,int y){
	return ((fact[y]*inv[x])%mod*inv[y-x])%mod;
}
signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	int n,m,i,j,k,l,ans=0;
	cin>>n>>m;
	fact[0]=1;
	for(i=1;i<=2*(n+m);i++){
		fact[i]=(fact[i-1]*i)%mod;
	}
	inv[2*(m+n)]=binpow(fact[2*(m+n)],mod-2);
	for(i=2*(m+n);i>=1;i--){
		inv[i-1]=(inv[i]*i)%mod;
	}
	for(i=m%2;i<=min(n,m);i+=2){
		if(i>3*m){
			break;
		}
		ans+=C(i,n)*C(n-1,(3*m-i)/2+n-1);
		if(i<=m){
			ans-=(n*C(i,n))%mod*C(n-1,n-1+(m-i)/2);
			if(i<n){
				ans+=(n*C(i,n-1))%mod*C(n-2,n-2+(m-i)/2);
			}
		}
		ans%=mod;
		if(ans<0){
			ans+=mod;
		}
	}
	cout<<ans;
}
### Prompt

Develop a solution in cpp to the problem described below:

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>
#define int long long
using namespace std;
const int mod=998244353;
const int N=3e6+2;
int fact[N],inv[N];
int binpow(int x,int y){
	int tich=1;
	while(y){
		if((y&1)){
			tich*=x;
			tich%=mod;
		}
		x*=x;
		x%=mod;
		y>>=1;
	}
	return tich;
}
int C(int x,int y){
	return ((fact[y]*inv[x])%mod*inv[y-x])%mod;
}
signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	int n,m,i,j,k,l,ans=0;
	cin>>n>>m;
	fact[0]=1;
	for(i=1;i<=2*(n+m);i++){
		fact[i]=(fact[i-1]*i)%mod;
	}
	inv[2*(m+n)]=binpow(fact[2*(m+n)],mod-2);
	for(i=2*(m+n);i>=1;i--){
		inv[i-1]=(inv[i]*i)%mod;
	}
	for(i=m%2;i<=min(n,m);i+=2){
		if(i>3*m){
			break;
		}
		ans+=C(i,n)*C(n-1,(3*m-i)/2+n-1);
		if(i<=m){
			ans-=(n*C(i,n))%mod*C(n-1,n-1+(m-i)/2);
			if(i<n){
				ans+=(n*C(i,n-1))%mod*C(n-2,n-2+(m-i)/2);
			}
		}
		ans%=mod;
		if(ans<0){
			ans+=mod;
		}
	}
	cout<<ans;
}
```