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>

#define PI 3.14159265358979323846264338327950288419

using namespace std;

const int mod = 998244353;
int n, m;
int fact[2000005], inv_fact[2000005];

int expo(int a, int b){
	int c = 1, d = a;
	while(b){
		if(b & 1){
			c = 1LL * c * d % mod;
		}
		d = 1LL * d * d % mod;
		b >>= 1;
	}
	return c;
}

int inv(int a){
	return expo(a, mod - 2);
}

int ncr(int _n, int _r){
	if(_n < 0 || _r < 0 || _r > _n){
		return 0;
	}
	int ret = fact[_n];
	ret = 1LL * ret * inv_fact[_r] % mod;
	ret = 1LL * ret * inv_fact[_n - _r] % mod;
	return ret;
}
	

int main(){
	fact[0] = inv_fact[0] = 1;
	for(int i = 1; i <= 2000000; i++){
		fact[i] = 1LL * fact[i - 1] * i % mod;
		inv_fact[i] = inv(fact[i]);
	}
	
	scanf("%d %d", &n, &m);
	int ans = 0;
	for(int odd = m; odd >= 0; odd -= 2){
		// ways to assign odds?
		if(odd > n){
			continue;
		}
		int combin_pt1 = ncr(n, odd);
		int remain = (3 * m - odd) / 2;
		
		// first assume we assign randomly
		int combin_pt2 = ncr(remain + n - 1, n - 1);
		
		// fail case 1: an odd number gets >= m out of remain
		combin_pt2 = (combin_pt2 - 1LL * odd * ncr(remain - m + n - 1, n - 1)) % mod;
		
		// fail case 2: an even number gets > m + 1 out of remaining
		combin_pt2 = (combin_pt2 - 1LL * (n - odd) * ncr(remain - (m + 1) + n - 1, n - 1)) % mod;
		
		ans = (ans + 1LL * combin_pt1 * combin_pt2) % mod;
	}
	if(ans < 0){
		ans += mod;
	}
	printf("%d\n", ans);
	return 0;
}