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;

const int N = 3000300;
const int mod = 998244353;

void add(int &x, int y) {
	x += y;
	if (x >= mod) x -= mod;
}

void sub(int &x, int y) {
	x -= y;
	if (x < 0) x += mod;
}

int mul(int x, int y) {
	return (long long) x * y % mod;
}

int pw(int x, int y) {
	int ans = 1;
	for (; y > 0; y >>= 1, x = mul(x, x)) {
		if (y & 1) ans = mul(ans, x);
	} 
	return ans;
}

int fact[N];
int ifact[N];

int C(int n, int k) {
	if (n < k || k < 0) return 0;
	return mul(fact[n], mul(ifact[k], ifact[n - k]));
}

void precalc() {
	fact[0] = 1;
	for (int i = 1; i < N; ++i) {
		fact[i] = mul(fact[i - 1], i);
	}		
	ifact[N - 1] = pw(fact[N - 1], mod - 2);
	for (int i = N - 1; i > 0; --i) {
		ifact[i - 1] = mul(ifact[i], i);
	}
}

int main() {
	ios_base::sync_with_stdio(false);
	precalc();
	int n, m;
	cin >> n >> m;
	int ans = C(3 * m + n - 1, n - 1);
	sub(ans, mul(n, C(m - 1 + n - 1, n - 1)));
	for (int i = m + 2; i <= n; i += 2) {
		sub(ans, mul(C(n, i), C((3 * m - i) / 2 + n - 1, n - 1)));
	}
	cout << ans << "\n";
}