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 <bits/stdc++.h>
using namespace std;

typedef long long ll;

const int mod = 998244353;
const int Maxn = 3000005;

int fac[Maxn], inv[Maxn];
int n, m;
int res;

int toPower(int a, int p)
{
    int res = 1;
    while (p) {
        if (p & 1) res = ll(res) * a % mod;
        p >>= 1; a = ll(a) * a % mod;
    }
    return res;
}

int Inv(int x) { return toPower(x, mod - 2); }

int C(int n, int k)
{
    if (n < 0 || k < 0 || k > n) return 0;
    return ll(fac[n]) * inv[k] % mod * inv[n - k] % mod;
}

int main()
{
    fac[0] = inv[0] = 1;
    for (int i = 1; i < Maxn; i++) {
        fac[i] = ll(i) * fac[i - 1] % mod;
        inv[i] = Inv(fac[i]);
    }
    scanf("%d %d", &n, &m);
    for (int x = 0; x <= n && x <= m; x++) if ((3 * m - x) % 2 == 0) {
        int S = (3 * m - x) / 2;
        // take m
        res = (res + ll(n) * C(S - m + n - 2, n - 2) % mod * C(n - 1, x)) % mod;
        // not take m
        int av = m;
        int ways = (C(S + n - 1, n - 1) - ll(n) * C(S - av + n - 1, n - 1) % mod + mod) % mod;
        res = (res + ll(ways) * C(n, x)) % mod;
    }
    printf("%d\n", res);
    return 0;
}
