#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<queue>
#include<string.h>
using namespace std;
typedef long long ll;
const ll M=998244353;
int n,i,j,m;
ll p[5000005],inv[5000005],ans;
ll C(ll n,ll m)
{
	if(n<m)
		return 0;
	return p[n]*inv[m]%M*inv[n-m]%M;
}
ll qpow(ll a,ll b)
{
	ll s=a,ans=1;
	while(b)
	{
		if(b&1)
			ans=ans*s%M;
		s=s*s%M;
		b>>=1;
	}
	return ans;
}
int main()
{
	scanf("%d %d",&n,&m);
	p[0]=p[1]=inv[1]=1;
	for(i=2;i<=m*3+n;i++)
	{
		p[i]=p[i-1]*i%M;
		inv[i]=(M-M/i)*inv[M%i]%M;
	}
	inv[0]=1;
	for(i=2;i<=m*3+n;i++)
		inv[i]=inv[i]*inv[i-1]%M;
	ans=C(m*3+n-1,n-1)-C(m+n-2,n-1)*n%M;
	for(i=m+1;i<=m*3&&i<=n;i++)
		if((m*3-i)%2==0)
			ans=(ans-C(n,i)*C(n+(m*3-i)/2-1,n-1))%M;
	cout<<(ans%M+M)%M;
}

### 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<stdio.h>
#include<iostream>
#include<algorithm>
#include<queue>
#include<string.h>
using namespace std;
typedef long long ll;
const ll M=998244353;
int n,i,j,m;
ll p[5000005],inv[5000005],ans;
ll C(ll n,ll m)
{
	if(n<m)
		return 0;
	return p[n]*inv[m]%M*inv[n-m]%M;
}
ll qpow(ll a,ll b)
{
	ll s=a,ans=1;
	while(b)
	{
		if(b&1)
			ans=ans*s%M;
		s=s*s%M;
		b>>=1;
	}
	return ans;
}
int main()
{
	scanf("%d %d",&n,&m);
	p[0]=p[1]=inv[1]=1;
	for(i=2;i<=m*3+n;i++)
	{
		p[i]=p[i-1]*i%M;
		inv[i]=(M-M/i)*inv[M%i]%M;
	}
	inv[0]=1;
	for(i=2;i<=m*3+n;i++)
		inv[i]=inv[i]*inv[i-1]%M;
	ans=C(m*3+n-1,n-1)-C(m+n-2,n-1)*n%M;
	for(i=m+1;i<=m*3&&i<=n;i++)
		if((m*3-i)%2==0)
			ans=(ans-C(n,i)*C(n+(m*3-i)/2-1,n-1))%M;
	cout<<(ans%M+M)%M;
}

```