#include <bits/stdc++.h>
#define mod 1000000007
#define mod998 998244353
#define sp ' '
#define intmax 2147483647
#define llmax 9223372036854775807
#define mkp make_pair
typedef long long ll;
using namespace std;

int N, M;
ll f[2000001], res;

ll pw(ll x, int y) {
	ll a = 1;
	while (y) {
		if (y & 1)a = a*x%mod998;
		x = x*x%mod998;
		y /= 2;
	}
	return a;
}

ll modinv(ll x) {
	return pw(x, mod998 - 2);
}

ll comb(int n, int r) {
	if (n < r)return 0;
	return f[n] * modinv(f[r] * f[n - r] % mod998) % mod998;
}

int main() {
	cin >> N >> M;
	f[0] = 1;
	for (int i = 1; i <= 2000000; ++i) {
		f[i] = f[i - 1] * i%mod998;
	}
	for (int i = 0; i <= min(N, M); ++i) {
		if ((3 * M - i) % 2)continue;
		int t = M + (M - i) / 2;
		res += (comb(t + N - 1, N - 1) + mod998 - comb(t - M + N - 1, N - 1)*i%mod998 + mod998 - comb(t - M - 1 + N - 1, N - 1)*(N - i) % mod998) % mod998*comb(N, i);
		res %= mod998;
	}
	cout << res << endl;
}
### Prompt

Please create a solution in Cpp 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>
#define mod 1000000007
#define mod998 998244353
#define sp ' '
#define intmax 2147483647
#define llmax 9223372036854775807
#define mkp make_pair
typedef long long ll;
using namespace std;

int N, M;
ll f[2000001], res;

ll pw(ll x, int y) {
	ll a = 1;
	while (y) {
		if (y & 1)a = a*x%mod998;
		x = x*x%mod998;
		y /= 2;
	}
	return a;
}

ll modinv(ll x) {
	return pw(x, mod998 - 2);
}

ll comb(int n, int r) {
	if (n < r)return 0;
	return f[n] * modinv(f[r] * f[n - r] % mod998) % mod998;
}

int main() {
	cin >> N >> M;
	f[0] = 1;
	for (int i = 1; i <= 2000000; ++i) {
		f[i] = f[i - 1] * i%mod998;
	}
	for (int i = 0; i <= min(N, M); ++i) {
		if ((3 * M - i) % 2)continue;
		int t = M + (M - i) / 2;
		res += (comb(t + N - 1, N - 1) + mod998 - comb(t - M + N - 1, N - 1)*i%mod998 + mod998 - comb(t - M - 1 + N - 1, N - 1)*(N - i) % mod998) % mod998*comb(N, i);
		res %= mod998;
	}
	cout << res << endl;
}
```