#include<iostream>
using namespace std;
long N,M,mod=998244353;
long F[1<<25],I[1<<25];
long power(long a,long b){return b?power(a*a%mod,b/2)*(b%2?a:1)%mod:1;}
long C(long a,long b)
{
	return a<b||a<0||b<0?0:F[a]*I[b]%mod*I[a-b]%mod;
}
long H(long a,long b){return a<0?0:C(a+b-1,b-1);}
long calc(long n,long m,long s)
{
	long ret=0;
	for(int i=0;i<=n&&i<=m;i++)
	{
		if(s-i&1)continue;
		(ret+=C(n,i)*H((s-i)/2,n)%mod)%=mod;
	}
	return ret;
}
int main()
{
	cin>>N>>M;
	long T=2*M+N;
	F[0]=1;
	for(long i=1;i<T;i++)F[i]=F[i-1]*i%mod;
	I[T-1]=power(F[T-1],mod-2);
	for(long i=T-1;i--;)I[i]=I[i+1]*(i+1)%mod;
	long ans=0;
	for(int i=0;i<=N&&i<=M;i++)
	{
		if((3*M-i)%2==1)continue;
		(ans+=C(N,i)*H((3*M-i)/2,N)%mod)%=mod;
	}
	long ret=0;
	for(int i=0;i<=N&&i<=M;i++)
	{
		if((M-i)%2==1)continue;
		(ret+=C(N,i)*H((M-i)/2,N)%mod)%=mod;
	}
	for(int i=0;i<=N-1&&i<=M;i++)
	{
		if((M-i)%2==1)continue;
		(ret+=mod-C(N-1,i)*H((M-i)/2,N-1)%mod)%=mod;
	}
	cout<<(calc(N,M,3*M)-calc(N,M,M)*N%mod+calc(N-1,M,M)*N%mod+mod)%mod<<endl;
}

### Prompt

In Cpp, your task is to solve 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>
using namespace std;
long N,M,mod=998244353;
long F[1<<25],I[1<<25];
long power(long a,long b){return b?power(a*a%mod,b/2)*(b%2?a:1)%mod:1;}
long C(long a,long b)
{
	return a<b||a<0||b<0?0:F[a]*I[b]%mod*I[a-b]%mod;
}
long H(long a,long b){return a<0?0:C(a+b-1,b-1);}
long calc(long n,long m,long s)
{
	long ret=0;
	for(int i=0;i<=n&&i<=m;i++)
	{
		if(s-i&1)continue;
		(ret+=C(n,i)*H((s-i)/2,n)%mod)%=mod;
	}
	return ret;
}
int main()
{
	cin>>N>>M;
	long T=2*M+N;
	F[0]=1;
	for(long i=1;i<T;i++)F[i]=F[i-1]*i%mod;
	I[T-1]=power(F[T-1],mod-2);
	for(long i=T-1;i--;)I[i]=I[i+1]*(i+1)%mod;
	long ans=0;
	for(int i=0;i<=N&&i<=M;i++)
	{
		if((3*M-i)%2==1)continue;
		(ans+=C(N,i)*H((3*M-i)/2,N)%mod)%=mod;
	}
	long ret=0;
	for(int i=0;i<=N&&i<=M;i++)
	{
		if((M-i)%2==1)continue;
		(ret+=C(N,i)*H((M-i)/2,N)%mod)%=mod;
	}
	for(int i=0;i<=N-1&&i<=M;i++)
	{
		if((M-i)%2==1)continue;
		(ret+=mod-C(N-1,i)*H((M-i)/2,N-1)%mod)%=mod;
	}
	cout<<(calc(N,M,3*M)-calc(N,M,M)*N%mod+calc(N-1,M,M)*N%mod+mod)%mod<<endl;
}

```