#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
#define mod 998244353
#define mad(a,b) a=(a+b)%mod
#define mul(a,b) a=(a*b)%mod
#define N 5000010
ll fac[N],inv[N],ivf[N];
ll C(ll n,ll k){
    if(n<k)return 0;
    return fac[n]*ivf[n-k]%mod*ivf[k]%mod;
}
ll H(ll n,ll k){
    //if(k==0)return 1;
    return C(n+k-1,k);
}
ll P(ll n,ll k){
    return fac[n]*ivf[n-k]%mod;
}
ll po(ll x,ll y){
    ll res=1;
    for(;y;y>>=1){
	if(y&1)res=res*x%mod;
	x=x*x%mod;
    }
    return res;
}

int main(){
    fac[0]=ivf[0]=inv[1]=1;
    for(int i=1;i<N;i++){
	if(i>1)inv[i]=(mod-mod/i*inv[mod%i]%mod)%mod;
	fac[i]=fac[i-1]*i%mod;
	ivf[i]=ivf[i-1]*inv[i]%mod;
    }
     ll n,m; cin>>n>>m;
     ll ans=0;
     for(ll p=0;p<=m;p++){
	 if((3*m-p)%2==1)continue;
	 ll cur=H(n,(3*m-p)/2)*C(n,p);
	 mad(ans,cur);
	 //cout<<cur<<endl;
     }
     mad(ans,-n*H(n,m-1));
     if(ans<0)ans+=mod;
     cout<<ans<<endl;
    return 0;
}

### Prompt

Please create a solution in Cpp to 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<iostream>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
#define mod 998244353
#define mad(a,b) a=(a+b)%mod
#define mul(a,b) a=(a*b)%mod
#define N 5000010
ll fac[N],inv[N],ivf[N];
ll C(ll n,ll k){
    if(n<k)return 0;
    return fac[n]*ivf[n-k]%mod*ivf[k]%mod;
}
ll H(ll n,ll k){
    //if(k==0)return 1;
    return C(n+k-1,k);
}
ll P(ll n,ll k){
    return fac[n]*ivf[n-k]%mod;
}
ll po(ll x,ll y){
    ll res=1;
    for(;y;y>>=1){
	if(y&1)res=res*x%mod;
	x=x*x%mod;
    }
    return res;
}

int main(){
    fac[0]=ivf[0]=inv[1]=1;
    for(int i=1;i<N;i++){
	if(i>1)inv[i]=(mod-mod/i*inv[mod%i]%mod)%mod;
	fac[i]=fac[i-1]*i%mod;
	ivf[i]=ivf[i-1]*inv[i]%mod;
    }
     ll n,m; cin>>n>>m;
     ll ans=0;
     for(ll p=0;p<=m;p++){
	 if((3*m-p)%2==1)continue;
	 ll cur=H(n,(3*m-p)/2)*C(n,p);
	 mad(ans,cur);
	 //cout<<cur<<endl;
     }
     mad(ans,-n*H(n,m-1));
     if(ans<0)ans+=mod;
     cout<<ans<<endl;
    return 0;
}

```