#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>

typedef long long LL;

const int N=3000005;
const int MOD=998244353;

int n,m,jc[N],ny[N];

int C(int n,int m)
{
	if (n<m) return 0;
	else return (LL)jc[n]*ny[m]%MOD*ny[n-m]%MOD;
}

LL solve(int x,int y)
{
	int ans=0;
	for (int i=0;i*m<=x;i++)
		if (i&1) (ans+=MOD-(LL)C(y,i)*C(x-i*m+y-1,y-1)%MOD)%=MOD;
		else (ans+=(LL)C(y,i)*C(x-i*m+y-1,y-1)%MOD)%=MOD;
	return ans;
}

int main()
{
	scanf("%d%d",&n,&m);
	jc[0]=jc[1]=ny[0]=ny[1]=1;
	for (int i=2;i<=3000000;i++) jc[i]=(LL)jc[i-1]*i%MOD,ny[i]=(LL)(MOD-MOD/i)*ny[MOD%i]%MOD;
	for (int i=2;i<=3000000;i++) ny[i]=(LL)ny[i]*ny[i-1]%MOD;
	int ans=0;
	for (int i=0;i<=std::min(n,m);i++)
	{
		if ((i&1)!=(m&1)) continue;
		(ans+=(LL)solve((m*3-i)/2,n)*C(n,i)%MOD)%=MOD;
		if (m-i>=0) (ans+=(LL)solve((m-i)/2,n-1)*n%MOD*C(n-1,i)%MOD)%=MOD;
	}
	printf("%d\n",ans);
	return 0;
}
### Prompt

Create a solution in CPP for 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>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>

typedef long long LL;

const int N=3000005;
const int MOD=998244353;

int n,m,jc[N],ny[N];

int C(int n,int m)
{
	if (n<m) return 0;
	else return (LL)jc[n]*ny[m]%MOD*ny[n-m]%MOD;
}

LL solve(int x,int y)
{
	int ans=0;
	for (int i=0;i*m<=x;i++)
		if (i&1) (ans+=MOD-(LL)C(y,i)*C(x-i*m+y-1,y-1)%MOD)%=MOD;
		else (ans+=(LL)C(y,i)*C(x-i*m+y-1,y-1)%MOD)%=MOD;
	return ans;
}

int main()
{
	scanf("%d%d",&n,&m);
	jc[0]=jc[1]=ny[0]=ny[1]=1;
	for (int i=2;i<=3000000;i++) jc[i]=(LL)jc[i-1]*i%MOD,ny[i]=(LL)(MOD-MOD/i)*ny[MOD%i]%MOD;
	for (int i=2;i<=3000000;i++) ny[i]=(LL)ny[i]*ny[i-1]%MOD;
	int ans=0;
	for (int i=0;i<=std::min(n,m);i++)
	{
		if ((i&1)!=(m&1)) continue;
		(ans+=(LL)solve((m*3-i)/2,n)*C(n,i)%MOD)%=MOD;
		if (m-i>=0) (ans+=(LL)solve((m-i)/2,n-1)*n%MOD*C(n-1,i)%MOD)%=MOD;
	}
	printf("%d\n",ans);
	return 0;
}
```