#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

//  a^b % m
long long powmod(long long a, long long b, long long m)
{
    long long r = 1;
    for (; b>0; b>>=1, a=a*a%m)
        if (b&1)
            r = r*a%m;
    return r;
}

int main()
{
    long long N, M;
    cin>>N>>M;

    long long MOD = 998244353;
    
    vector<long long> F(M*3/2+N);
    F[0] = 1;
    for (int i=1; i<M*3/2+N; i++)
        F[i] = F[i-1]*i%MOD;

    auto comb = [&](long long a, long long b)
    {
        return F[a]*powmod(F[b], MOD-2, MOD)%MOD*powmod(F[a-b], MOD-2, MOD)%MOD;
    };

    auto f = [&](long long m)
    {
        long long ans = 0;
        for (int o=m%2; o<=min(M, N); o+=2)
            ans += comb(N, o)*comb((m-o)/2+N-1, N-1)%MOD;
        return ans%MOD;
    };

    long long ans = ((f(3*M)-N*f(M-1))%MOD+MOD)%MOD;
    cout<<ans<<endl;
}

### 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 <iostream>
#include <vector>
#include <algorithm>
using namespace std;

//  a^b % m
long long powmod(long long a, long long b, long long m)
{
    long long r = 1;
    for (; b>0; b>>=1, a=a*a%m)
        if (b&1)
            r = r*a%m;
    return r;
}

int main()
{
    long long N, M;
    cin>>N>>M;

    long long MOD = 998244353;
    
    vector<long long> F(M*3/2+N);
    F[0] = 1;
    for (int i=1; i<M*3/2+N; i++)
        F[i] = F[i-1]*i%MOD;

    auto comb = [&](long long a, long long b)
    {
        return F[a]*powmod(F[b], MOD-2, MOD)%MOD*powmod(F[a-b], MOD-2, MOD)%MOD;
    };

    auto f = [&](long long m)
    {
        long long ans = 0;
        for (int o=m%2; o<=min(M, N); o+=2)
            ans += comb(N, o)*comb((m-o)/2+N-1, N-1)%MOD;
        return ans%MOD;
    };

    long long ans = ((f(3*M)-N*f(M-1))%MOD+MOD)%MOD;
    cout<<ans<<endl;
}

```