#include<bits/stdc++.h>
#define LL long long
#define fr(i,x,y) for(int i=(x);i<=(y);i++)
#define rf(i,x,y) for(int i=(x);i>=(y);i--)
#define frl(i,x,y) for(int i=(x);i<(y);i++)
using namespace std;
const int N=3000002;
const int p=998244353;
int n,m;
LL mul[N],inv[N];
LL ans;

void read(int &x){ scanf("%d",&x); }
void read(LL &x){ scanf("%lld",&x); }

LL qpow(LL a,int n){
	LL ans=1;
	for(LL sum=a;n;n>>=1,sum=sum*sum%p) if (n&1) ans=ans*sum%p;
	return ans;
}

void init(){
	mul[0]=1;
	frl(i,1,N) mul[i]=mul[i-1]*i%p;
	inv[N-1]=qpow(mul[N-1],p-2);
	rf(i,N-2,0) inv[i]=inv[i+1]*(i+1)%p;
}

LL C(int n,int m){
	if (n<0||m<0||n-m<0) return 0;
	return mul[n]*inv[m]%p*inv[n-m]%p;
}

void Add(LL &x,LL y){
	x+=y;
	while(x<0) x+=p;
	while(x>=p) x-=p;
}

int main(){
	read(n);read(m);
	init();
	fr(i,0,m)
	 if ((3*m-i)%2==0) Add(ans,C(n,i)*C((3*m-i)/2+n-1,n-1)%p);
	fr(i,0,m)
	 if ((m-i)%2==0) Add(ans,-C(n,i)*C((m-i)/2+n-1,n-1)%p*n%p);
	fr(i,0,m)
	 if ((m-i)%2==0) Add(ans,C(n-1,i)*C((m-i)/2+n-2,n-2)%p*n%p);
	cout<<ans<<endl;
	return 0;
}
### Prompt

In Cpp, your task is to solve 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 LL long long
#define fr(i,x,y) for(int i=(x);i<=(y);i++)
#define rf(i,x,y) for(int i=(x);i>=(y);i--)
#define frl(i,x,y) for(int i=(x);i<(y);i++)
using namespace std;
const int N=3000002;
const int p=998244353;
int n,m;
LL mul[N],inv[N];
LL ans;

void read(int &x){ scanf("%d",&x); }
void read(LL &x){ scanf("%lld",&x); }

LL qpow(LL a,int n){
	LL ans=1;
	for(LL sum=a;n;n>>=1,sum=sum*sum%p) if (n&1) ans=ans*sum%p;
	return ans;
}

void init(){
	mul[0]=1;
	frl(i,1,N) mul[i]=mul[i-1]*i%p;
	inv[N-1]=qpow(mul[N-1],p-2);
	rf(i,N-2,0) inv[i]=inv[i+1]*(i+1)%p;
}

LL C(int n,int m){
	if (n<0||m<0||n-m<0) return 0;
	return mul[n]*inv[m]%p*inv[n-m]%p;
}

void Add(LL &x,LL y){
	x+=y;
	while(x<0) x+=p;
	while(x>=p) x-=p;
}

int main(){
	read(n);read(m);
	init();
	fr(i,0,m)
	 if ((3*m-i)%2==0) Add(ans,C(n,i)*C((3*m-i)/2+n-1,n-1)%p);
	fr(i,0,m)
	 if ((m-i)%2==0) Add(ans,-C(n,i)*C((m-i)/2+n-1,n-1)%p*n%p);
	fr(i,0,m)
	 if ((m-i)%2==0) Add(ans,C(n-1,i)*C((m-i)/2+n-2,n-2)%p*n%p);
	cout<<ans<<endl;
	return 0;
}
```