#include<bits/stdc++.h>
using namespace std;

const int _ = 3e6 + 7 , MOD = 998244353;
int jc[_] , inv[_] , N , M;

int poww(long long a , int b){
	int times = 1;
	while(b){if(b & 1) times = times * a % MOD; a = a * a % MOD; b >>= 1;}
	return times;
}

void init(int mx){
	jc[0] = 1; for(int i = 1 ; i <= mx ; ++i) jc[i] = 1ll * jc[i - 1] * i % MOD;
	inv[mx] = poww(jc[mx] , MOD - 2);
	for(int i = mx - 1 ; i >= 0 ; --i) inv[i] = inv[i + 1] * (i + 1ll) % MOD;
}

int binom(int p , int q){return p < q ? 0 : 1ll * jc[p] * inv[q] % MOD * inv[p - q] % MOD;}

int main(){
	cin >> N >> M; init(3 * M + N); int sum = 0;
	for(int i = 0 ; i <= N && i <= M ; ++i){
		if((3 * M - i) & 1) continue;
		sum = (sum + 1ll * binom(N , i) * binom((3 * M - i) / 2 + N - 1 , N - 1)) % MOD;
	}
	cout << (sum - 1ll * N * binom(M - 1 + N - 1 , N - 1) % MOD + MOD) % MOD; return 0;
}

### Prompt

Your task is to create a CPP solution to the following 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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;

const int _ = 3e6 + 7 , MOD = 998244353;
int jc[_] , inv[_] , N , M;

int poww(long long a , int b){
	int times = 1;
	while(b){if(b & 1) times = times * a % MOD; a = a * a % MOD; b >>= 1;}
	return times;
}

void init(int mx){
	jc[0] = 1; for(int i = 1 ; i <= mx ; ++i) jc[i] = 1ll * jc[i - 1] * i % MOD;
	inv[mx] = poww(jc[mx] , MOD - 2);
	for(int i = mx - 1 ; i >= 0 ; --i) inv[i] = inv[i + 1] * (i + 1ll) % MOD;
}

int binom(int p , int q){return p < q ? 0 : 1ll * jc[p] * inv[q] % MOD * inv[p - q] % MOD;}

int main(){
	cin >> N >> M; init(3 * M + N); int sum = 0;
	for(int i = 0 ; i <= N && i <= M ; ++i){
		if((3 * M - i) & 1) continue;
		sum = (sum + 1ll * binom(N , i) * binom((3 * M - i) / 2 + N - 1 , N - 1)) % MOD;
	}
	cout << (sum - 1ll * N * binom(M - 1 + N - 1 , N - 1) % MOD + MOD) % MOD; return 0;
}

```