#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <cmath>
#include <map>
#include <queue>
#include <iomanip>
#include <set>
#define MOD 998244353
#define mkp make_pair
typedef long long ll;
using namespace std;

#define MAX_N 4000040
ll inv[MAX_N+10],fac[MAX_N+10],ifac[MAX_N+10];

void setComb(){
  inv[0]=1;inv[1]=1;fac[1]=1;ifac[1]=1;fac[0]=1;ifac[0]=1;
  for(int i=2;i<MAX_N;i++){
    inv[i]=(-MOD/i)*inv[MOD%i]%MOD;
    fac[i]=fac[i-1]*i%MOD;
    ifac[i]=ifac[i-1]*inv[i]%MOD;

    inv[i]=(inv[i]+MOD)%MOD;
    fac[i]=(fac[i]+MOD)%MOD;
    ifac[i]=(ifac[i]+MOD)%MOD;
  }
  return;
}

ll comb(ll n,ll k){
  if(n<k||n<0||k<0) return 0;
  else return ((fac[n]*ifac[k]%MOD*ifac[n-k]%MOD+MOD)%MOD);
}

ll hcomb(ll n,ll r){
  if(n==0&&r==0) return 1;
  else if(n<0||r<0) return 0;
  else return comb(n+r-1,r);
}

ll mod_pow(ll x,ll n){
  x%=MOD;
  ll res=1;
  while(n>0){
    if(n&1) res=res*x%MOD;
    x=x*x%MOD;
    n>>=1;
  }
  return res;
}

void add(ll &a,ll b){
  a=(a+b)%MOD;
}

void mul(ll &a,ll b){
  a%=MOD;b%=MOD;
  a=a*b%MOD;
}

int main(){
  ll N,M;
  cin>>N>>M;

  setComb();

  ll nans=0;
  for(ll k=0;k<M;k++){
    ll res=hcomb(N-1,k);
    mul(res,N);
    add(nans,res);
  }

  ll ans=0;
  for(ll k=0;k<=min(N,M);k++){
    if(k%2!=M%2) continue;
    ll res=hcomb(N,(3*M-k)/2);
    mul(res,comb(N,k));
    add(ans,res);
  }

  add(ans,-nans+MOD);
  cout<<ans<<endl;

  return 0;
}

### Prompt

Create a solution in Cpp for 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 <string>
#include <vector>
#include <cmath>
#include <map>
#include <queue>
#include <iomanip>
#include <set>
#define MOD 998244353
#define mkp make_pair
typedef long long ll;
using namespace std;

#define MAX_N 4000040
ll inv[MAX_N+10],fac[MAX_N+10],ifac[MAX_N+10];

void setComb(){
  inv[0]=1;inv[1]=1;fac[1]=1;ifac[1]=1;fac[0]=1;ifac[0]=1;
  for(int i=2;i<MAX_N;i++){
    inv[i]=(-MOD/i)*inv[MOD%i]%MOD;
    fac[i]=fac[i-1]*i%MOD;
    ifac[i]=ifac[i-1]*inv[i]%MOD;

    inv[i]=(inv[i]+MOD)%MOD;
    fac[i]=(fac[i]+MOD)%MOD;
    ifac[i]=(ifac[i]+MOD)%MOD;
  }
  return;
}

ll comb(ll n,ll k){
  if(n<k||n<0||k<0) return 0;
  else return ((fac[n]*ifac[k]%MOD*ifac[n-k]%MOD+MOD)%MOD);
}

ll hcomb(ll n,ll r){
  if(n==0&&r==0) return 1;
  else if(n<0||r<0) return 0;
  else return comb(n+r-1,r);
}

ll mod_pow(ll x,ll n){
  x%=MOD;
  ll res=1;
  while(n>0){
    if(n&1) res=res*x%MOD;
    x=x*x%MOD;
    n>>=1;
  }
  return res;
}

void add(ll &a,ll b){
  a=(a+b)%MOD;
}

void mul(ll &a,ll b){
  a%=MOD;b%=MOD;
  a=a*b%MOD;
}

int main(){
  ll N,M;
  cin>>N>>M;

  setComb();

  ll nans=0;
  for(ll k=0;k<M;k++){
    ll res=hcomb(N-1,k);
    mul(res,N);
    add(nans,res);
  }

  ll ans=0;
  for(ll k=0;k<=min(N,M);k++){
    if(k%2!=M%2) continue;
    ll res=hcomb(N,(3*M-k)/2);
    mul(res,comb(N,k));
    add(ans,res);
  }

  add(ans,-nans+MOD);
  cout<<ans<<endl;

  return 0;
}

```