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 <bits/stdc++.h>
using namespace std;;
#define ll long long
#define rep(i, n) for(int i = 0;i < n;i++)
#define repr(i, n) for(int i = n;i >= 0;i--)
#define FOR(i, m, n) for(ll i = m;i < n;i++)
#define FORR(i, m, n) for(int i = m;i >= n;i--)
#define INF 1<<30
#define LINF 1LL<<62
#define all(x) (x).begin(), (x).end()
#define mp make_pair
#define pb push_back
const ll MOD = 998244353;
 
typedef pair<ll, ll> P;
typedef pair<ll, P> PP;

const int MAX = 3000000;
long long fac[MAX], finv[MAX], inv[MAX];
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
int main(){
  ios::sync_with_stdio(false);
  cin.tie(0);
  COMinit();
  ll n,m;cin >>n >> m;
  ll ans = COM(3*m + n - 1 , n-1);
  FOR(i,2*m+1, 3*m+1){
      ll x = 3*m - i;
      ll tmp = n * COM(x + n - 2 , n - 2) % MOD;
      ans += MOD;
      ans -= tmp ;
      ans %= MOD;
  }
  FOR(i,m+1,n+1){
      if(3*m<i)break;
      if((3*m-i)%2==1)continue;
      ll eve = (3*m-i)/2;
      ll x = COM(n,i) * COM(eve + n - 1, n-1) % MOD;
      ans += MOD;
      ans -= x;
      ans %=MOD;
  }
  cout << ans << endl;
  
}