#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll p=998244353,jc[4000010],ny[4000010];
ll ksm(ll x,ll y){
	ll xlh=1;
	while(y){
		if(y&1)xlh=xlh*x%p;
		x=x*x%p;
		y/=2;
	}
	return xlh;
}
ll C(ll x,ll y){
	if(x<0||y<0||x<y)return 0;
	return jc[x]*ny[y]%p*ny[x-y]%p;
}
ll solve(ll n,ll zo,ll m){
	ll i,ans=0,xlh,sy;
	for(i=0;i<=m;i++)if(zo%2==i%2){
		xlh=C(n,i);
		sy=zo-i;sy/=2;
		xlh=xlh*C(sy-1+n,n-1)%p;
		ans=(ans+xlh)%p;
	}
	return ans;
}
int main(){
	ll i,n,m;
	jc[0]=ny[0]=1;
	for(i=1;i<=4000000;i++)jc[i]=jc[i-1]*i%p;
	ny[4000000]=ksm(jc[4000000],p-2);
	for(i=4000000-1;i;i--)ny[i]=ny[i+1]*(i+1)%p;
	scanf("%lld%lld",&n,&m);
	printf("%lld",(solve(n,m*3,m)-n*(solve(n,m,m)-solve(n-1,m,m)+p)%p+p)%p);
}
### Prompt

Please formulate 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 ll long long
using namespace std;
ll p=998244353,jc[4000010],ny[4000010];
ll ksm(ll x,ll y){
	ll xlh=1;
	while(y){
		if(y&1)xlh=xlh*x%p;
		x=x*x%p;
		y/=2;
	}
	return xlh;
}
ll C(ll x,ll y){
	if(x<0||y<0||x<y)return 0;
	return jc[x]*ny[y]%p*ny[x-y]%p;
}
ll solve(ll n,ll zo,ll m){
	ll i,ans=0,xlh,sy;
	for(i=0;i<=m;i++)if(zo%2==i%2){
		xlh=C(n,i);
		sy=zo-i;sy/=2;
		xlh=xlh*C(sy-1+n,n-1)%p;
		ans=(ans+xlh)%p;
	}
	return ans;
}
int main(){
	ll i,n,m;
	jc[0]=ny[0]=1;
	for(i=1;i<=4000000;i++)jc[i]=jc[i-1]*i%p;
	ny[4000000]=ksm(jc[4000000],p-2);
	for(i=4000000-1;i;i--)ny[i]=ny[i+1]*(i+1)%p;
	scanf("%lld%lld",&n,&m);
	printf("%lld",(solve(n,m*3,m)-n*(solve(n,m,m)-solve(n-1,m,m)+p)%p+p)%p);
}
```