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;

int main() {
    int n, m; scanf("%d%d", &n, &m);
    constexpr int kMod = 998244353;
    vector<int> fc(3 * m + n, 1), iv(3 * m + n, 1);

    auto fpow = [&](int a, int n) {
        int res = 1;
        while (n > 0) {
            if (n & 1) res = 1LL * res * a % kMod;
            a = 1LL * a * a % kMod;
            n >>= 1;
        }
        return res;
    };

    for (int i = 1; i < fc.size(); ++i) {
        fc[i] = 1LL * fc[i - 1] * i % kMod;
        iv[i] = fpow(fc[i], kMod - 2);
    }

    auto Choose = [&](int n, int k) {
        if (n < k || k < 0) return 0LL;
        return 1LL * fc[n] * iv[k] % kMod * iv[n - k] % kMod;
    };

    int ans = 0;
    for (int i = 0; i <= m; ++i) {
        if ((3 * m - i) % 2 != 0) continue;
        ans += 1LL * Choose((3 * m - i) / 2 + n - 1, n - 1) * Choose(n, i) % kMod;
        ans %= kMod;
    }

    vector<int> a(m), b(m);
    for (int i = 0; i < m; ++i) a[i] = Choose(n - 1, i);
    for (int i = 0; i < m; i += 2) b[i] = Choose(i / 2 + n - 2, n - 2);

    int sub = 0, y = 0;
    for (int i = m - 1, j = 0; i >= 0; --i) {
        while (j < m && i + j < m) (y += b[j++]) %= kMod;
        sub += 1LL * a[i] * y % kMod;
        if (sub >= kMod) sub -= kMod;
    }
    sub = 1LL * sub * n % kMod;
    printf("%d\n", (ans + kMod - sub) % kMod);
    return 0;
}