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

Solution (in CPP):

Solution:
#include <iostream>
#include <algorithm>
#include <map>
#include <vector>
using namespace std;
typedef long long int ll;

ll mod=998244353;

ll mod_pow(ll a,ll b){
	a%=mod;
	if(b==0)return 1;
	if(b==1)return a;
	ll res=mod_pow(a,b/2)%mod;
	res*=res; res%=mod;
	if(b%2)res*=a;
	return res%mod;
}

struct perm{
private:
	int sz;
	vector<ll> p,invp;
public:
	perm(int n){
		sz=n+1;
		p.resize(sz),invp.resize(sz);
		p[0]=1;
		for(int i=1;i<=sz-1;i++){
			p[i]=p[i-1]*i%mod;
		}
		invp[sz-1]=mod_pow(p[sz-1],mod-2);
		for(int i=sz-2;i>=0;i--){
			invp[i]=invp[i+1]*(i+1)%mod;
		}
	}
	ll comb(ll x,ll y){
		if(x<y||y<0)return 0;
		return (p[x]*invp[x-y]%mod)*invp[y]%mod;
	}
};
perm p(5000000);

int main(){
	cin.tie(nullptr);
	ios::sync_with_stdio(false);
	ll n,m; cin >> n >> m;
	ll ans=0;
	for(int i=0;i<=min(n,m);i++){
		if((3*m-i)%2)continue;
		ll k=(3*m-i)/2;
		(ans+=p.comb(n,i)*p.comb(n-1+k,k)%mod)%=mod;
	}
	ll res=0;
	ll res2=0;
	for(int i=0;i<=(m,n);i++){
		if((m-i)%2)continue;
		ll k=(m-i)/2;
		(res+=p.comb(n,i)*p.comb(n-1+k,k)%mod)%=mod;
	}
	for(int i=0;i<=(m,n-1);i++){
		if((m-i)%2)continue;
		ll k=(m-i)/2;
		(res2+=p.comb(n-1,i)*p.comb(n-2+k,k)%mod)%=mod;
	}
	res-=res2;
	res%=mod;
	ans-=res*n;
	ans%=mod;
	if(ans<0)ans+=mod;
	cout << ans << endl;
}