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 MOD = 998244353;

long long modpow(long long x, long long p){
	if(p == 0)return 1;
	if(p % 2 == 1)return x * modpow(x, p - 1) % MOD;
	
	long long r = modpow(x, p / 2);
	return r * r % MOD;
}
long long modinv(long long x){
	return modpow(x, MOD - 2);
}

int main(){
	int n, m;
	scanf("%d %d", &n, &m);
	
	int k = 3 * m + n;
	vector <long long> fact(k + 1), invfact(k + 1);
	
	fact[0] = 1;
	for(int i = 1; i <= k; i++)
		fact[i] = i * fact[i - 1] % MOD;
		
	invfact[k] = modinv(fact[k]);
	for(int i = k - 1; i >= 0; i--)
		invfact[i] = invfact[i + 1] * (i + 1) % MOD;

	auto C = [&](int n, int r) -> long long{
				return fact[n] * invfact[r] % MOD * invfact[n - r] % MOD;
			};
	
	long long res = 0;
	
	for(int odd = 0; odd <= min(n, m); odd++){
		if(odd % 2 != m % 2)continue;
		res = (res + C(n, odd) * C((3 * m - odd) / 2 + n - 1, n - 1)) % MOD;
	}

	for(int mx = 2 * m + 1; mx <= 3 * m; mx++)
		res = (res + (MOD - n) * C(3 * m - mx + n - 2, n - 2)) % MOD;
	
	printf("%lld\n", res);
	
	return 0;
}
