#include <bits/stdc++.h>
using namespace std;
const int mod=998244353;
const int N=4000010;
int i,j,k,n,m,x,y,t,ans;
int f[N],inv[N];
int mi(int x,int y){
	if (y==0)return 1;
	int t=mi(x,y>>1);t=1ll*t*t%mod;
	return y&1?(1ll*x*t%mod):t;
}
int C(int x,int y){return 1ll*f[x]*inv[y]%mod*inv[x-y]%mod;}
int main(){
	scanf("%d%d",&n,&m);
	f[0]=1;for (i=1;i<=4000000;i++)f[i]=1ll*i*f[i-1]%mod;
	inv[4000000]=mi(f[4000000],mod-2);for (i=3999999;i>=0;i--)inv[i]=1ll*inv[i+1]*(i+1)%mod;
	for (k=0;k<=min(n,m);k++)
		if ((3*m-k)%2==0){
			int w=(3*m-k)/2;
			ans=(ans+1ll*C(w+n-1,n-1)*C(n,k)%mod)%mod;
			if (w>=m){
				ans=(ans+mod-1ll*n*C(w-m+n-1,n-1)%mod*C(n,k)%mod)%mod;
				ans=(ans+1ll*n*C(w-m+n-2,n-2)%mod*C(n-1,k)%mod)%mod;
			}
		}
	printf("%d\n",ans);
	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>
using namespace std;
const int mod=998244353;
const int N=4000010;
int i,j,k,n,m,x,y,t,ans;
int f[N],inv[N];
int mi(int x,int y){
	if (y==0)return 1;
	int t=mi(x,y>>1);t=1ll*t*t%mod;
	return y&1?(1ll*x*t%mod):t;
}
int C(int x,int y){return 1ll*f[x]*inv[y]%mod*inv[x-y]%mod;}
int main(){
	scanf("%d%d",&n,&m);
	f[0]=1;for (i=1;i<=4000000;i++)f[i]=1ll*i*f[i-1]%mod;
	inv[4000000]=mi(f[4000000],mod-2);for (i=3999999;i>=0;i--)inv[i]=1ll*inv[i+1]*(i+1)%mod;
	for (k=0;k<=min(n,m);k++)
		if ((3*m-k)%2==0){
			int w=(3*m-k)/2;
			ans=(ans+1ll*C(w+n-1,n-1)*C(n,k)%mod)%mod;
			if (w>=m){
				ans=(ans+mod-1ll*n*C(w-m+n-1,n-1)%mod*C(n,k)%mod)%mod;
				ans=(ans+1ll*n*C(w-m+n-2,n-2)%mod*C(n-1,k)%mod)%mod;
			}
		}
	printf("%d\n",ans);
	return 0;
}
```