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
#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N=3000005,M=998244353;
int fac[N],inv[N],n,m,ans;
int ksm(int x,int y){
	int ans=1;
	for (;y;y>>=1,x=x*x%M)
		if (y&1)(ans*=x)%=M;
	return ans;	
}
int C(int x,int y){
	if (x<y)return 0;
	return fac[x]*inv[y]%M*inv[x-y]%M;
}
signed main(){
	scanf("%lld%lld",&n,&m);
	fac[0]=inv[0]=1;
	for (int i=1;i<N;i++)fac[i]=fac[i-1]*i%M,inv[i]=ksm(fac[i],M-2);
	for (int i=m&1;i<=m;i+=2)
		(ans+=C(n+(3*m-i)/2-1,n-1)*C(n,i))%=M;
	for (int i=m&1;i<=m;i+=2)
		(ans+=M-n*C(n+(m-i)/2-1,n-1)%M*C(n,i)%M)%=M;
	n--;
	for (int i=m&1;i<=m;i+=2)
		(ans+=(n+1)*C(n+(m-i)/2-1,n-1)%M*C(n,i)%M)%=M;		
	printf("%lld\n",ans);
	return 0;
} 