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<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#define mod 998244353
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0' && ch<='9')x=x*10+ch-'0',ch=getchar();
    return x*f;
}
inline void write(int x)
{
    if(x<0)putchar('-'),x=-x;
    if(x>9)write(x/10);
    putchar(x%10+'0');
}
inline void pr1(int x){write(x),putchar(' ');}
inline void pr2(int x){write(x),puts("");}
int fac[2500010],inv[2500010];
inline int power(int a,int b)
{
	int ans=1;
	while(b)
	{
		if(b&1)ans=1LL*ans*a%mod;
		a=1LL*a*a%mod;
		b>>=1;
	}return ans;
}
inline int C(int n,int m)
{
	if(n<0 || m<0)return 0;
	return 1LL*fac[n]*inv[m]%mod*inv[n-m]%mod;
}
int main()
{
    //freopen("a.in","r",stdin);
    //freopen("a.out","w",stdout);
    fac[0]=1;
    for(int i=1;i<=2500000;i++)fac[i]=1LL*fac[i-1]*i%mod;inv[2500000]=power(fac[2500000],mod-2);
    for(int i=2500000;i>=1;i--)inv[i-1]=1LL*inv[i]*i%mod;
    int n=read(),m=read(),ans;
    ans=C(n-1+3*m,n-1);
    for(int i=m+1;i<=min(n,3*m);i++)
	{
		if((3*m-i)&1)continue;
		(ans-=1LL*C(n,i)*C((3*m-i)/2+n-1,n-1)%mod-mod)%=mod;
	}
    for(int i=2*m+1;i<=3*m;i++)(ans-=1LL*n*C(3*m-i+n-2,n-2)%mod-mod)%=mod;
    pr2(ans);
    return 0;
}