#include<bits/stdc++.h>
using namespace std;
const int N=3e6+5,mod=998244353;
typedef long long ll;
int n,m;
ll f[N],invf[N];
ll C(ll n,ll m){return f[n]*invf[n-m]%mod*invf[m]%mod;}
int main()
{
    f[0]=f[1]=invf[0]=invf[1]=1;
    for(int i=2;i<N;i++)
    {
        f[i]=f[i-1]*i%mod;
        invf[i]=(mod-mod/i)*invf[mod%i]%mod;
    }
    for(int i=2;i<N;i++) invf[i]=invf[i-1]*invf[i]%mod;
    scanf("%d%d",&n,&m);
    ll ans=0,res=0;
    for(int i=0;i<=min(n,m);i++)
        if((3*m-i)%2==0)
        ans=(ans+C(n,i)*C((3*m-i)/2+n-1,n-1)%mod)%mod;
    for(int i=0;i<=min(n,m);i++)
        if((m-i)%2==0)
        res=(res+C(n,i)*C((m-i)/2+n-1,n-1)%mod)%mod;
    ans=(ans-res*n%mod)%mod;
    ans=(ans+mod)%mod;
    for(int i=0;i<=min(n-1,m);i++)
        if((m-i)%2==0)
        ans=(ans+C(n-1,i)*C((m-i)/2+n-2,n-2)%mod*n)%mod;
    printf("%lld\n",ans);
}

### Prompt

Your task is to create a CPP solution 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<bits/stdc++.h>
using namespace std;
const int N=3e6+5,mod=998244353;
typedef long long ll;
int n,m;
ll f[N],invf[N];
ll C(ll n,ll m){return f[n]*invf[n-m]%mod*invf[m]%mod;}
int main()
{
    f[0]=f[1]=invf[0]=invf[1]=1;
    for(int i=2;i<N;i++)
    {
        f[i]=f[i-1]*i%mod;
        invf[i]=(mod-mod/i)*invf[mod%i]%mod;
    }
    for(int i=2;i<N;i++) invf[i]=invf[i-1]*invf[i]%mod;
    scanf("%d%d",&n,&m);
    ll ans=0,res=0;
    for(int i=0;i<=min(n,m);i++)
        if((3*m-i)%2==0)
        ans=(ans+C(n,i)*C((3*m-i)/2+n-1,n-1)%mod)%mod;
    for(int i=0;i<=min(n,m);i++)
        if((m-i)%2==0)
        res=(res+C(n,i)*C((m-i)/2+n-1,n-1)%mod)%mod;
    ans=(ans-res*n%mod)%mod;
    ans=(ans+mod)%mod;
    for(int i=0;i<=min(n-1,m);i++)
        if((m-i)%2==0)
        ans=(ans+C(n-1,i)*C((m-i)/2+n-2,n-2)%mod*n)%mod;
    printf("%lld\n",ans);
}

```