#include <iostream>
#include <iomanip>
#include <cstdio>
#include <vector>
#include <string>
#include <algorithm>
#include <map>
#include <set>
 
using namespace std;
typedef long long ll;
const int MAXN = 4000100;
const ll MOD = 998244353;

int N, M;
ll ninv[MAXN];
ll fac[MAXN];
ll finv[MAXN];
ll npow[MAXN], mpow[MAXN];

ll rchoose (int x, int y)
{
    ll res = (fac[x] * finv[y]) % MOD;
    return (res * finv[x-y]) % MOD;
}

ll distro (int K)
{
    if (K < 0) return 0;
    return rchoose (N + K - 1, K);
}

int main()
{
    ios_base::sync_with_stdio(0);
    ninv[1] = 1;
    for (int i = 2; i < MAXN; i++)
    {
        int K = MOD % i;
        ll r = ((ninv[K] * (K - MOD)) / i) % MOD + MOD;
        ninv[i] = r % MOD;
    }

    fac[0] = finv[0] = npow[0] = mpow[0] = 1;
    for (int i = 1; i < MAXN; i++)
    {
        fac[i] = (fac[i-1] * i) % MOD;
        finv[i] = (finv[i-1] * ninv[i]) % MOD;
    }
    
    cin >> N >> M;

    ll ans = 0;
    for (int i = 0; i <= N; i++)
    {
        int K = i;
        if ((i + M) % 2 != 0) continue;
        if (i > M) continue;
        ll l = rchoose (N, i);
        ll r = (distro ((3 * M - K) / 2) - (K * distro ((M - K) / 2)) % MOD - ((N - K) * distro ((M - K) / 2 - 1)) % MOD) % MOD;
        ll v = (l * r) % MOD;
        //cout << K << " " << l << " " << r << " " << v << "\n";
        ans = (ans + v) % MOD;
    }
    cout << ((ans % MOD + MOD) % MOD) << "\n";
}
### Prompt

Construct a cpp code solution to the problem outlined:

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 <iostream>
#include <iomanip>
#include <cstdio>
#include <vector>
#include <string>
#include <algorithm>
#include <map>
#include <set>
 
using namespace std;
typedef long long ll;
const int MAXN = 4000100;
const ll MOD = 998244353;

int N, M;
ll ninv[MAXN];
ll fac[MAXN];
ll finv[MAXN];
ll npow[MAXN], mpow[MAXN];

ll rchoose (int x, int y)
{
    ll res = (fac[x] * finv[y]) % MOD;
    return (res * finv[x-y]) % MOD;
}

ll distro (int K)
{
    if (K < 0) return 0;
    return rchoose (N + K - 1, K);
}

int main()
{
    ios_base::sync_with_stdio(0);
    ninv[1] = 1;
    for (int i = 2; i < MAXN; i++)
    {
        int K = MOD % i;
        ll r = ((ninv[K] * (K - MOD)) / i) % MOD + MOD;
        ninv[i] = r % MOD;
    }

    fac[0] = finv[0] = npow[0] = mpow[0] = 1;
    for (int i = 1; i < MAXN; i++)
    {
        fac[i] = (fac[i-1] * i) % MOD;
        finv[i] = (finv[i-1] * ninv[i]) % MOD;
    }
    
    cin >> N >> M;

    ll ans = 0;
    for (int i = 0; i <= N; i++)
    {
        int K = i;
        if ((i + M) % 2 != 0) continue;
        if (i > M) continue;
        ll l = rchoose (N, i);
        ll r = (distro ((3 * M - K) / 2) - (K * distro ((M - K) / 2)) % MOD - ((N - K) * distro ((M - K) / 2 - 1)) % MOD) % MOD;
        ll v = (l * r) % MOD;
        //cout << K << " " << l << " " << r << " " << v << "\n";
        ans = (ans + v) % MOD;
    }
    cout << ((ans % MOD + MOD) % MOD) << "\n";
}
```