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 <string>
#include <cstdio>
#include <iostream>
#include <vector>
#include <unordered_map>
#include <cstring>
#include <algorithm>
#include <bitset>
#include <deque>
#include <stack>
#include <queue>
using namespace std;
const int N = 1e7;
const int mod = 998244353;
int n, m;
long long a[N +1], ni[N + 1];
long long calc(long long x, long long y) {
    long long z = 1;
    while (y) {
        if (y & 1) (z *= x )%= mod;
        (x *= x) %= mod, y /= 2;
    }
    return z;
}
long long C(int x, int y) {
    if (x < y || x < 0) return 0;
    return a[x] * ni[y] % mod * ni[x - y] % mod;
}
long long get(int n, int m) {
    return C(n + m  - 1, n - 1);
}
int main()
{
    scanf("%d %d", &n ,&m);
    a[0] = 1;
    for (int i = 1; i <= N ; i ++)
        a[i] = a[i - 1] * i % mod;
    ni[N] = calc(a[N] , mod -  2);
    for (int i = N - 1; i >= 0 ; i --)
         ni[i] = ni[i + 1] * (i + 1) % mod;
    long long ans = get(n, m * 3);
    for (int i = m + 1; i <= n ; i ++) {
        if ((3 * m - i) % 2 == 0 && m * 3 - i >= 0)
            (ans += mod - C(n, i) * get(n, (m * 3 - i) / 2) % mod) %= mod;
    }
    (ans += mod - C(n , 1) * get(n, m * 3 - m - m - 1)% mod) %= mod;
    printf("%d\n", ans);
    return 0;
}