#include <bits/stdc++.h>
using namespace std;
int N,M;
const long long fish=998244353;
const int MaxN=(1<<21)-1;
long long inv[MaxN+15],fac[MaxN+15];
long long Pow(int x,int y){
	long long b=x,ans=1;
	for (int qwq=y;qwq;qwq>>=1){
		if (qwq&1) ans=(ans*1ll*b)%fish;
		b=(b*b)%fish;
	}
	return ans;
}
void Init(){
	fac[0]=1;
	for (int i=1;i<=MaxN;i++)
		fac[i]=1ll*fac[i-1]*i%fish;
	inv[MaxN]=Pow(fac[MaxN],fish-2);
	for (int i=MaxN-1;i>=0;i--)
		inv[i]=1ll*inv[i+1]*(i+1)%fish;
	//cout<<inv[2]<<endl;
}
long long C(int n,int r){
	return fac[n]*1ll*inv[r]%fish*1ll*inv[n-r]%fish;
}
long long Get_ans(int n,int a,int b){
	b=min(b,n);
	long long ans=0;
	for (int i=a%2;i<=min(min(a,b),n);i+=2)
		ans+=1ll*C((a-i)/2+n-1,n-1)%fish*C(n,i)%fish;
		//cout<<ans<<endl;
		//cout<<n<<endl;
	return ans;
}
int main(){
	Init();
	scanf("%d%d",&N,&M);
	//cout<<C(3,2)<<endl;
	long long ans=Get_ans(N,3*M,M);
	//cout<<ans<<endl;
	ans-=(Get_ans(N,M,M)-Get_ans(N-1,M,M))*1ll*N%fish;
	cout<<(ans+fish)%fish;
	return 0;
}
### Prompt

Your challenge is to write 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 <bits/stdc++.h>
using namespace std;
int N,M;
const long long fish=998244353;
const int MaxN=(1<<21)-1;
long long inv[MaxN+15],fac[MaxN+15];
long long Pow(int x,int y){
	long long b=x,ans=1;
	for (int qwq=y;qwq;qwq>>=1){
		if (qwq&1) ans=(ans*1ll*b)%fish;
		b=(b*b)%fish;
	}
	return ans;
}
void Init(){
	fac[0]=1;
	for (int i=1;i<=MaxN;i++)
		fac[i]=1ll*fac[i-1]*i%fish;
	inv[MaxN]=Pow(fac[MaxN],fish-2);
	for (int i=MaxN-1;i>=0;i--)
		inv[i]=1ll*inv[i+1]*(i+1)%fish;
	//cout<<inv[2]<<endl;
}
long long C(int n,int r){
	return fac[n]*1ll*inv[r]%fish*1ll*inv[n-r]%fish;
}
long long Get_ans(int n,int a,int b){
	b=min(b,n);
	long long ans=0;
	for (int i=a%2;i<=min(min(a,b),n);i+=2)
		ans+=1ll*C((a-i)/2+n-1,n-1)%fish*C(n,i)%fish;
		//cout<<ans<<endl;
		//cout<<n<<endl;
	return ans;
}
int main(){
	Init();
	scanf("%d%d",&N,&M);
	//cout<<C(3,2)<<endl;
	long long ans=Get_ans(N,3*M,M);
	//cout<<ans<<endl;
	ans-=(Get_ans(N,M,M)-Get_ans(N-1,M,M))*1ll*N%fish;
	cout<<(ans+fish)%fish;
	return 0;
}
```