#include<bits/stdc++.h>
#define ll long long
#define mk(x,y) make_pair(x,y)
#define pb(x) push_back(x)
using namespace std;
const int mod=998244353;
const int N=2e6+10;
int inv[N],fac[N];
int n,m,Ans;
int mul(int x,int y){return (ll)x*y%mod;}
int add(int x,int y){return x+y>=mod?x+y-mod:x+y;}
int dec(int x,int y){return x-y<0?x-y+mod:x-y;}
int ksm(int x,int y){
	int ans=1;
	for (;y;y>>=1,x=mul(x,x)) if (y&1) ans=mul(ans,x);
	return ans;
}
void init(){
	fac[0]=1;
	for (int i=1;i<N;i++) fac[i]=mul(fac[i-1],i);
	inv[0]=inv[1]=1;
	for (int i=2;i<N;i++) inv[i]=mul(mod-mod/i,inv[mod%i]);
	for (int i=2;i<N;i++) inv[i]=mul(inv[i],inv[i-1]);
}
int C(int x,int y){
	if (x<y||x<0||y<0) return 0;
	return mul(mul(fac[x],inv[y]),inv[x-y]);
}
int main(){
	scanf("%d%d",&n,&m);
	Ans=0; init();
	for (int i=0;i<=min(n,m);i++) 
	if ((m*3-i)%2==0){
		int x=(m*3-i)/2;
		int s=C(n+x-1,n-1);
		s=dec(s,mul(n,C(n+x-m-1,n-1)));
		s=add(s,mul(n-i,C(n+x-m-2,n-2)));
		s=mul(s,C(n,i));
		Ans=add(Ans,s);
	}
	printf("%d\n",Ans);
}
### 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 mk(x,y) make_pair(x,y)
#define pb(x) push_back(x)
using namespace std;
const int mod=998244353;
const int N=2e6+10;
int inv[N],fac[N];
int n,m,Ans;
int mul(int x,int y){return (ll)x*y%mod;}
int add(int x,int y){return x+y>=mod?x+y-mod:x+y;}
int dec(int x,int y){return x-y<0?x-y+mod:x-y;}
int ksm(int x,int y){
	int ans=1;
	for (;y;y>>=1,x=mul(x,x)) if (y&1) ans=mul(ans,x);
	return ans;
}
void init(){
	fac[0]=1;
	for (int i=1;i<N;i++) fac[i]=mul(fac[i-1],i);
	inv[0]=inv[1]=1;
	for (int i=2;i<N;i++) inv[i]=mul(mod-mod/i,inv[mod%i]);
	for (int i=2;i<N;i++) inv[i]=mul(inv[i],inv[i-1]);
}
int C(int x,int y){
	if (x<y||x<0||y<0) return 0;
	return mul(mul(fac[x],inv[y]),inv[x-y]);
}
int main(){
	scanf("%d%d",&n,&m);
	Ans=0; init();
	for (int i=0;i<=min(n,m);i++) 
	if ((m*3-i)%2==0){
		int x=(m*3-i)/2;
		int s=C(n+x-1,n-1);
		s=dec(s,mul(n,C(n+x-m-1,n-1)));
		s=add(s,mul(n-i,C(n+x-m-2,n-2)));
		s=mul(s,C(n,i));
		Ans=add(Ans,s);
	}
	printf("%d\n",Ans);
}
```