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;
typedef long long ll;

const int MAX = 2510000;
const int mod = 998244353;

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

// テーブルを作る前処理
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;
    }
}

// 二項係数計算
ll 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;
}

signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << fixed << setprecision(20);
    
    COMinit();
    ll n,m;
    cin>>n>>m;
    ll ans = COM(3*m+n-1,n-1);
    // cerr << ans << endl;
    for(int k=2*m+1;k<=3*m;k++){
        ll ret = n*COM(3*m-k+n-2,n-2);
        ret %= mod;
        // cerr << k << " " << ret << endl;
        ans += mod - ret;
        ans %= mod;
    }
    int x = 1;
    for(int i=m+1;i<=n;i++){
        ll ret = COM(n,i);
        if((m*3-i)%2)continue;
        ll p = 3*m-i;
        p /= 2;
        ret *= COM(n+p-1,n-1);
        ret %= mod;
        ans += mod - ret;
        ans %= mod;
    }
    ans %= mod;
    cout << ans << endl;
    
}