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>

using namespace std;


typedef long long ll;

const long MOD = 998244353;
#define N_MAX 2000002

ll inv[N_MAX],fac[N_MAX],finv[N_MAX];

void init(){
    fac[0]=fac[1]=1;
    finv[0]=finv[1]=1;
    inv[1]=1;
    for(int i=2;i<N_MAX;i++){
        inv[i]=MOD-inv[MOD%i]*(MOD/i)%MOD;
        fac[i]=fac[i-1]*(ll) i%MOD;
        finv[i]=finv[i-1]*inv[i]%MOD;
    }
}

ll inv_(ll n){
    if(n == 1) return 1;
    else return MOD-inv_(MOD%n)*(MOD/n)%MOD;
}

ll comb(ll n, ll r){
  ll ans;
  if(n < r){
      ans = 0;
  }else{
      ans = (fac[n]*finv[r])%MOD;
      ans = (ans*finv[n-r])%MOD;
      ans = (ans+MOD)%MOD;
  }
  return ans;
}

int main(){
    ll N, M;
    cin >> N >> M;
    init();
    ll ans = 0;
    for(ll m = 0; m <= M; m++){
        if((M-m)%2 != 0) continue;
        ans += (comb(N, m)*comb((3*M-m)/2+N-1, N-1))%MOD;
        ans %= MOD;
        ans -= (((comb(N, m)*(N-m))%MOD)*comb((M-m)/2-1+N-1, N-1))%MOD;
        ans %= MOD;
        ans += MOD;
        ans %= MOD;
        ans -= (((comb(N, m)*m)%MOD)*comb((M-m)/2+N-1, N-1))%MOD;
        ans %= MOD;
        ans += MOD;
        ans %= MOD;
        //cout << ans << endl;
    }
    cout << ans << endl;
}