#include<bits/stdc++.h>
using namespace std;

const int mod = 998244353;
long long int fac[5000001],inv[5000001],cnt[5000001];

long long int C(int n,int k){
	return fac[n]*inv[k]%mod*inv[n-k]%mod;
}

long long int solve(int n,int m,int k){
	long long int res = 0;
	for(int i=0;i<=k;i++)
		if(n-i>=0&&(n-i)%2==0&&m>=i)
			res += C((n-i)/2+m-1,m-1) * C(m,i) % mod;
	return res % mod;
}

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	fac[0]=fac[1]=inv[0]=inv[1]=1;
	for(int i=2;i<=5000000;i++){
		fac[i]=fac[i-1]*i%mod;
		inv[i]=mod-mod/i*inv[mod%i]%mod;
	}
	for(int i=2;i<=1000000;i++)
		inv[i]=inv[i-1]*inv[i]%mod;
	int n,m;
	cin>>n>>m;
	long long int ans = solve(3*m,n,m) - (solve(m,n,m) - solve(m,n-1,m)) * n;
	cout<<(ans%mod+mod)%mod<<endl;
}

### Prompt

Please create a solution in CPP 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;

const int mod = 998244353;
long long int fac[5000001],inv[5000001],cnt[5000001];

long long int C(int n,int k){
	return fac[n]*inv[k]%mod*inv[n-k]%mod;
}

long long int solve(int n,int m,int k){
	long long int res = 0;
	for(int i=0;i<=k;i++)
		if(n-i>=0&&(n-i)%2==0&&m>=i)
			res += C((n-i)/2+m-1,m-1) * C(m,i) % mod;
	return res % mod;
}

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	fac[0]=fac[1]=inv[0]=inv[1]=1;
	for(int i=2;i<=5000000;i++){
		fac[i]=fac[i-1]*i%mod;
		inv[i]=mod-mod/i*inv[mod%i]%mod;
	}
	for(int i=2;i<=1000000;i++)
		inv[i]=inv[i-1]*inv[i]%mod;
	int n,m;
	cin>>n>>m;
	long long int ans = solve(3*m,n,m) - (solve(m,n,m) - solve(m,n-1,m)) * n;
	cout<<(ans%mod+mod)%mod<<endl;
}

```