#include<bits/stdc++.h>
using namespace std;
const int maxn=4001000,mod=998244353;
inline void Add(int &a,int b){a=a+b>=mod?a+b-mod:a+b;}
inline int ksm(int a,int b)
{
	int res=1;
	for(;b;b>>=1,a=1ll*a*a%mod) if(b&1) res=1ll*res*a%mod;
	return res;
}
int fac[maxn],invfac[maxn];
inline int C(int n,int m){return 1ll*fac[n]*invfac[m]%mod*invfac[n-m]%mod;}
int n,m;
inline int cal(int n,int m,int lim)
{
	int ans=0;
	for(int i=0;i<=min(n,lim);i++) if(!((m-i)&1))
		Add(ans,1ll*C(n,i)*C((m-i)/2+n-1,n-1)%mod);
	return ans;
}
int main()
{
	for(int i=fac[0]=1;i<maxn;i++) fac[i]=1ll*fac[i-1]*i%mod;
	invfac[maxn-1]=ksm(fac[maxn-1],mod-2);
	for(int i=maxn-2;i>=0;i--) invfac[i]=1ll*invfac[i+1]*(i+1)%mod;
	cin>>n>>m;printf("%lld\n",(cal(n,3*m,m)-1ll*n*cal(n,m,m)%mod+1ll*n*cal(n-1,m,m)%mod+mod)%mod);
	return 0;
}
### Prompt

Generate 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<bits/stdc++.h>
using namespace std;
const int maxn=4001000,mod=998244353;
inline void Add(int &a,int b){a=a+b>=mod?a+b-mod:a+b;}
inline int ksm(int a,int b)
{
	int res=1;
	for(;b;b>>=1,a=1ll*a*a%mod) if(b&1) res=1ll*res*a%mod;
	return res;
}
int fac[maxn],invfac[maxn];
inline int C(int n,int m){return 1ll*fac[n]*invfac[m]%mod*invfac[n-m]%mod;}
int n,m;
inline int cal(int n,int m,int lim)
{
	int ans=0;
	for(int i=0;i<=min(n,lim);i++) if(!((m-i)&1))
		Add(ans,1ll*C(n,i)*C((m-i)/2+n-1,n-1)%mod);
	return ans;
}
int main()
{
	for(int i=fac[0]=1;i<maxn;i++) fac[i]=1ll*fac[i-1]*i%mod;
	invfac[maxn-1]=ksm(fac[maxn-1],mod-2);
	for(int i=maxn-2;i>=0;i--) invfac[i]=1ll*invfac[i+1]*(i+1)%mod;
	cin>>n>>m;printf("%lld\n",(cal(n,3*m,m)-1ll*n*cal(n,m,m)%mod+1ll*n*cal(n-1,m,m)%mod+mod)%mod);
	return 0;
}
```