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<cstdio>
const int MOD = 998244353;
const int MAXN = 2000000;
int fct[MAXN + 5], ifct[MAXN + 5];
int pow_mod(int b, int p) {
	int ret = 1;
	while( p ) {
		if( p & 1 ) ret = 1LL*ret*b%MOD;
		b = 1LL*b*b%MOD;
		p >>= 1;
	}
	return ret;
}
void init() {
	fct[0] = 1;
	for(int i=1;i<=MAXN;i++)
		fct[i] = 1LL*fct[i-1]*i%MOD;
	ifct[MAXN] = pow_mod(fct[MAXN], MOD-2);
	for(int j=MAXN-1;j>=0;j--)
		ifct[j] = 1LL*ifct[j+1]*(j+1)%MOD;
}
int comb(int n, int m) {
	return 1LL*fct[n]*ifct[m]%MOD*ifct[n-m]%MOD;
}
int main() {
	init();
	int ans = 0, N, M;
	scanf("%d%d", &N, &M);
	for(int i=M&1;i<=M&&i<=N;i+=2) {
		int tmp = comb((3*M-i)/2+(N-1), N-1);
		tmp = (tmp + MOD - 1LL*(N-i)*comb((M-i-2)/2+(N-1), N-1)%MOD)%MOD;
		tmp = (tmp + MOD - 1LL*i*comb((M-i)/2+(N-1), N-1)%MOD)%MOD;
		ans = (ans + 1LL*tmp*comb(N, i)%MOD)%MOD;
	}
	printf("%d\n", ans);
}