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
#include <iostream>
using namespace std;
class ModComb {
    long long *fact, *facti;
    const int mod;
public:
    explicit ModComb(int n, int m) : mod(m) {
        fact = new long long[n+1];
        facti = new long long[n+1];
        fact[0] = 1; facti[0] = 1;
        for (int i = 1; i <= n; i++) fact[i] = (fact[i-1] * i) % m;
        // calc 1/n!
        long long &inv = facti[n], pw = fact[n];
        inv = 1;
        for (int e = mod-2; e > 0; e /= 2) {
            if (e&1) inv = inv * pw % mod;
            pw = pw * pw % mod;
        }
        for (int i = n-1; i > 0; i--) facti[i] = (facti[i+1] * (i+1)) % m;
    }

    ~ModComb() {
        if (fact) delete[] fact;
        if (facti) delete[] facti;
    }

    long long get(int n, int k) const {
        if (n < 0 || k < 0 || k > n) return 0;
        return fact[n] * facti[k] % mod * facti[n-k] % mod;
    }
};

int main() {
    int n, m; cin >> n >> m;
    const int M = 998244353;
    ModComb mc(3*m+n+3, M);

    long long ans = mc.get(n+3*m-1, n-1);
    for (int j = 0; j < m; j++) ans -= mc.get(n+j-2, n-2) * n % M;
    for (int k = m + 2; k <= n; k += 2)
        ans -= mc.get(n, k) * mc.get((3*m-k)/2+n-1, n-1) % M;
    cout << (ans % M + M) % M << endl;
}
