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>
 
#define endl '\n'
#define fi first
#define se second
#define MOD(n,k) ( ( ((n) % (k)) + (k) ) % (k))
#define forn(i,n) for (int i = 0; i < n; i++)
#define forr(i,a,b) for (int i = a; i <= b; i++)
#define all(v) v.begin(), v.end()
#define pb emplace_back
 
using namespace std;
 
typedef long long ll;
typedef long double ld;
typedef pair<int, int> ii;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<ii> vii;

const int MX = 5000005, mod = 998244353;
int n, m;
ll f[MX], in[MX];

ll comb (int n, int k) {
	return f[n] * in[k] % mod * in[n - k] % mod; 
}

ll fun (int n, int k) {
	return comb(n + k - 1, k);
}

int main () {
	ios_base::sync_with_stdio(0); cin.tie(0);
	f[0] = in[0] = in[1] = 1;
	for (int i = 1; i < MX; i++)
		f[i] = f[i-1] * i % mod;
	for (int i = 2; i < MX; i++)
		in[i] = in[mod % i] * (mod - mod / i) % mod;
	for (int i = 1; i < MX; i++)
		(in[i] *= in[i-1]) %= mod;

	cin >> n >> m;

	ll res = 0;
	for (int i = 0; i <= min(m, n); i++) {
		if ((3 * m - i) % 2) continue;
		(res += comb(n, i) * fun(n, (3 * m - i) / 2)) %= mod;
		(res -= comb(n, i) * fun(n, (m - i) / 2) % mod * n) %= mod;
		(res += comb(n, i) * fun(n - 1, (m - i) / 2) % mod * (n - i)) %= mod;
	}

	cout << MOD(res, mod) << endl;

	return 0;
}