#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define REP(i,n) for(int i=(0);i<(n);i++)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
using namespace std;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef long long ll;

const int N=3000005,mod=998244353;
int fac[N],rev[N],n,m,ans,sum;

inline int power(int x,int p){
	int res=1;
	for(;p;p>>=1,x=(ll)x*x%mod)
		if(p&1)res=(ll)res*x%mod;
	return res;
}
inline int C(int n,int m){
	return m<0||m>n?0:(ll)fac[n]*rev[m]%mod*rev[n-m]%mod;
}
void init(int lim){
	fac[0]=1;
	rep(i,1,lim)fac[i]=(ll)fac[i-1]*i%mod;
	rev[lim]=power(fac[lim],mod-2);
	per(i,lim,1)rev[i-1]=(ll)rev[i]*i%mod;
}

int calc(int n,int m){
	if(m<0||n<0)return 0;
	if(m==0&&n==0)return 1;
	return C(m+n-1,n-1);
}

int doit(int k){
	int rem=(sum-k)/2,mx=m;
	int sum=(calc(n,rem)-(ll)(n-k)*calc(n,rem-mx-1)%mod
			+2ll*mod-(ll)k*calc(n,rem-mx)%mod)%mod;
	//cerr<<k<<" "<<sum<<"  "<<rem<<endl;
	return (ll)C(n,k)*sum%mod;
}

int main(){
	scanf("%d%d",&n,&m);
	init(max(n,m)*3);
	sum=3*m;
	rep(i,0,min(m,n))
		if((sum-i)%2==0)
			ans=(ans+doit(i))%mod;
	printf("%d\n",ans);

	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>
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define REP(i,n) for(int i=(0);i<(n);i++)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
using namespace std;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef long long ll;

const int N=3000005,mod=998244353;
int fac[N],rev[N],n,m,ans,sum;

inline int power(int x,int p){
	int res=1;
	for(;p;p>>=1,x=(ll)x*x%mod)
		if(p&1)res=(ll)res*x%mod;
	return res;
}
inline int C(int n,int m){
	return m<0||m>n?0:(ll)fac[n]*rev[m]%mod*rev[n-m]%mod;
}
void init(int lim){
	fac[0]=1;
	rep(i,1,lim)fac[i]=(ll)fac[i-1]*i%mod;
	rev[lim]=power(fac[lim],mod-2);
	per(i,lim,1)rev[i-1]=(ll)rev[i]*i%mod;
}

int calc(int n,int m){
	if(m<0||n<0)return 0;
	if(m==0&&n==0)return 1;
	return C(m+n-1,n-1);
}

int doit(int k){
	int rem=(sum-k)/2,mx=m;
	int sum=(calc(n,rem)-(ll)(n-k)*calc(n,rem-mx-1)%mod
			+2ll*mod-(ll)k*calc(n,rem-mx)%mod)%mod;
	//cerr<<k<<" "<<sum<<"  "<<rem<<endl;
	return (ll)C(n,k)*sum%mod;
}

int main(){
	scanf("%d%d",&n,&m);
	init(max(n,m)*3);
	sum=3*m;
	rep(i,0,min(m,n))
		if((sum-i)%2==0)
			ans=(ans+doit(i))%mod;
	printf("%d\n",ans);

	return 0;
}
```