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>
using namespace std;

const int maxN = 3e6 + 10, lim = 3e6, mod = 998244353;

int n, m;
int fac[maxN + 1], inv[maxN + 1];
int ans;

inline int ADD(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }

inline int SUB(int x, int y) { return x - y < 0 ? x - y + mod : x - y; }

inline int mpow(int a, int x)
{
	int ans = 1;
	while(x)
	{
		if(x & 1) ans = 1ll * ans * a % mod;
		a = 1ll * a * a % mod;
		x >>= 1;
	}
	return ans;
}

inline int C(int n, int m)
{
	if(n < m) return 0;
	return 1ll * fac[n] * inv[m] % mod * inv[n - m] % mod;
}

int main()
{
	scanf("%d %d", &n, &m);

	fac[0] = 1;
	for(int i = 1; i <= lim; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
	inv[lim] = mpow(fac[lim], mod - 2);
	for(int i = lim - 1; i >= 0; i--) inv[i] = 1ll * inv[i + 1] * (i + 1) % mod;

	for(int i = 0; i <= min(n, m); i++)
	{
		if((3 * m - i) & 1) continue;
		int t = (3 * m - i) >> 1;
		ans = ADD(ans, 1ll * C(n, i) * C(t + n - 1, n - 1) % mod);
	}

	for(int i = 0; i <= min(n, m); i++)
	{
		if((3 * m - i) & 1) continue;
		int t = (3 * m - i) >> 1;
		if(i < n) ans = SUB(ans, 1ll * n * C(n - 1, i) % mod * C(t + n - m - 2, n - 1) % mod);
		if(i > 0) ans = SUB(ans, 1ll * n * C(n - 1, i - 1) % mod * C(t + n - m - 1, n - 1) % mod);
	}

	printf("%d\n", ans);
	return 0;
}