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;

#define int long long
#define double long double
#define mod 998244353ll
#define INF 1000000000000000000ll

#define MAX_N 5000010

int silnia[MAX_N];

void init() {
    silnia[0] = 1;
    for (int i = 1; i < MAX_N; i++) {
        silnia[i] = silnia[i - 1] * i % mod;
    }
}

int pot(int n, int k = mod - 2) {
    int res = 1;

    while (k) {
        if (k % 2) {
            res = res * n % mod;
        }
        n = n * n % mod;
        k /= 2;
    }

    return res;
}

int dwumian(int n, int k) {
    if (k > n || k < 0) return 0;
    return (silnia[n] * pot(silnia[n - k]) % mod) * pot(silnia[k]) % mod;
}

int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    init();

    int res = 0;
    int n, m;
    cin >> n >> m;
    //cout << dwumian(n, m);
    for (int i = m; 2 * i <= 3 * m; i++) {
        int a = dwumian(n - 1 + i, i);
        res += a * dwumian(n, 3 * m - 2 * i);
        res %= mod;
        //cout << a * dwumian(n, 3 * m - 2 * i) << " " << i  << endl;
    }

    for (int i = 2 * m + 1; i <= 3 * m; i++) {
        int a = n * dwumian((n - 1) + (3 * m - i) - 1, (n - 1) - 1) % mod;
        res = (res + mod - a) % mod;
    }

    cout << res;

    return 0;
}
/*

*/
