//by yjz
#include<bits/stdc++.h>
#define FF first
#define SS second
#define MP make_pair
#define PB push_back
typedef long long ll;
using namespace std;
const int mod = 998244353;
const int maxn = 5000111;
ll fac[maxn], inv[maxn], ifac[maxn];
ll C(int x, int y) {return fac[x]*ifac[y]%mod*ifac[x-y]%mod;}
ll F(int n, int m)
{
	ll ans = C(3*m+n-1, n-1);
	ll del = 0;
/*	for (int i=0; i<m; i++)
	{
		del = (del+C(i+n-2, n-2))%mod;
	}
*/
	del = C(m+n-2, n-1);
	ans = (ans-1ll*del*n)%mod;
	return (ans+mod)%mod;
}
ll S(int x, int y)
{
	if (y==0) return x==0;
	else
	{
		return C(x+y-1, y-1);
	}
}
int main()
{
	fac[0] = 1;
	for (int i=1; i<maxn; i++) fac[i] = fac[i-1]*i%mod;
	inv[1] = 1;
	for (int i=2; i<maxn; i++) inv[i] = mod-1ll*(mod/i)*inv[mod%i]%mod;
	ifac[0] = 1;
	for (int i=1; i<maxn; i++) ifac[i] = ifac[i-1]*inv[i]%mod;
	int n, m;
	cin>>n>>m;
	ll ans = F(n, m);
	for (int i=m+2; i<=3*m&&i<=n; i+=2)
	{
		ans = (ans-C(n, i)*S((3*m-i)/2, n))%mod; 
	}
	cout<<(ans+mod)%mod<<endl;
	return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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
//by yjz
#include<bits/stdc++.h>
#define FF first
#define SS second
#define MP make_pair
#define PB push_back
typedef long long ll;
using namespace std;
const int mod = 998244353;
const int maxn = 5000111;
ll fac[maxn], inv[maxn], ifac[maxn];
ll C(int x, int y) {return fac[x]*ifac[y]%mod*ifac[x-y]%mod;}
ll F(int n, int m)
{
	ll ans = C(3*m+n-1, n-1);
	ll del = 0;
/*	for (int i=0; i<m; i++)
	{
		del = (del+C(i+n-2, n-2))%mod;
	}
*/
	del = C(m+n-2, n-1);
	ans = (ans-1ll*del*n)%mod;
	return (ans+mod)%mod;
}
ll S(int x, int y)
{
	if (y==0) return x==0;
	else
	{
		return C(x+y-1, y-1);
	}
}
int main()
{
	fac[0] = 1;
	for (int i=1; i<maxn; i++) fac[i] = fac[i-1]*i%mod;
	inv[1] = 1;
	for (int i=2; i<maxn; i++) inv[i] = mod-1ll*(mod/i)*inv[mod%i]%mod;
	ifac[0] = 1;
	for (int i=1; i<maxn; i++) ifac[i] = ifac[i-1]*inv[i]%mod;
	int n, m;
	cin>>n>>m;
	ll ans = F(n, m);
	for (int i=m+2; i<=3*m&&i<=n; i+=2)
	{
		ans = (ans-C(n, i)*S((3*m-i)/2, n))%mod; 
	}
	cout<<(ans+mod)%mod<<endl;
	return 0;
}
```