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>
#define pb push_back
#define all(x) x.begin(), x.end()
using namespace std;
using ll = long long;
using vi = vector<ll>;
using vvi = vector<vi>;
using pi = pair<ll, ll>;
using vpi = vector<pi>;
const ll mod = 998244353;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int maxn = 4000000;
ll n, m, fact[maxn + 300], inv[maxn + 300];
ll bp(ll a, ll p) {
	ll res = 1;
	while (p) {
		if (p & 1)
			res = (res * a) % mod;
		a = (a * a) % mod;
		p >>= 1;
	}
	return res;
}
void setup() {
	fact[0] = 1;
	for (int i = 1; i <= maxn; i++)
		fact[i] = (fact[i - 1] * i) % mod;
	inv[maxn] = bp(fact[maxn], mod - 2);
	for (int i = maxn; i--;)
		inv[i] = (inv[i + 1] * (i + 1)) % mod;
}
ll nck(ll n, ll k) {
	ll t = (fact[n] * inv[n - k]) % mod;
	return (t * inv[k]) % mod;
}
ll solve(ll S, ll N) {
	return nck(S + N - 1, N - 1);
}
#define M(x) ((x)%mod)
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	setup();
	cin >> n >> m;
	ll ans = 0;
	for (int odd = 0; odd <= min(n, m); odd++) {
		if ((3 * m - odd) % 2 == 0)
			ans = (ans + solve((3 * m - odd) / 2, n) * nck(n, odd)) % mod;
	}
	for (int odd = 0; odd <= min(n, m); odd++) {
		if ((m - odd) % 2 == 0)
			ans = (mod + ans - M(n*M((solve((m - odd) / 2, n) * nck(n, odd))))%mod) % mod;
	}
	for (int odd = 0; odd <= min(n-1, m); odd++) {
		if ((m - odd) % 2 == 0)
			ans = (ans + M(n*M(solve((m - odd) / 2, n-1) * nck(n-1, odd)))) % mod;
	}
	cout << ans;
	return 0;
}
