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

const int maxn = 1750000, P = 998244353;
int n, m, fact[maxn], inv[maxn];

int qp(int x, int y) {
    int z = 1;
    for (; y; y >>= 1, x = 1LL * x * x % P) {
        if (y & 1) z = 1LL * z * x % P;
    }
    return z;
}

int C(int x, int y) {
    return 1LL * fact[x] * inv[x - y] % P * inv[y] % P;
}

int S(int s) {
    int ans = 0;
    for (int i = s & 1; i <= min({n, m, s}); i += 2) {
        ans = (ans + 1LL * C(n, i) * C((s - i) / 2 + n - 1, n - 1)) % P;
    }
    return ans;
}

int main() {
    for (int i = fact[0] = inv[0] = 1; i < maxn; i++) {
        inv[i] = qp(fact[i] = 1LL * i * fact[i - 1] % P, P - 2);
    }
    scanf("%d %d", &n, &m);
    printf("%d\n", (S(3 * m) - 1LL * n * S(m - 1) % P + P) % P);
    return 0;
}
### 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>
using namespace std;

const int maxn = 1750000, P = 998244353;
int n, m, fact[maxn], inv[maxn];

int qp(int x, int y) {
    int z = 1;
    for (; y; y >>= 1, x = 1LL * x * x % P) {
        if (y & 1) z = 1LL * z * x % P;
    }
    return z;
}

int C(int x, int y) {
    return 1LL * fact[x] * inv[x - y] % P * inv[y] % P;
}

int S(int s) {
    int ans = 0;
    for (int i = s & 1; i <= min({n, m, s}); i += 2) {
        ans = (ans + 1LL * C(n, i) * C((s - i) / 2 + n - 1, n - 1)) % P;
    }
    return ans;
}

int main() {
    for (int i = fact[0] = inv[0] = 1; i < maxn; i++) {
        inv[i] = qp(fact[i] = 1LL * i * fact[i - 1] % P, P - 2);
    }
    scanf("%d %d", &n, &m);
    printf("%d\n", (S(3 * m) - 1LL * n * S(m - 1) % P + P) % P);
    return 0;
}
```