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

ll modpower(ll a, ll r, ll mod){ //a^r
    ll x = 1;
    a %= mod;
    while (r > 0) {
        if (r & 1) x = x * a % mod;
        r >>= 1;
        a = a * a % mod;
    }
    return x;
}

ll modinv(ll a, ll mod){
    return modpower(a, mod-2, mod);
}

vector<ll> fact;
vector<ll> invfact;

void set_fact(ll n, ll mod){
    fact.resize(n+1, 1);
    invfact.resize(n+1, 1);
    for (ll i = 2; i <= n; i++) {
        fact[i] = fact[i-1] * i % mod;
    }
    invfact[n] = modinv(fact[n], mod);
    for (ll i = n-1; i >= 2; i--) {
        invfact[i] = invfact[i+1] * (i+1) % mod;
    }
    return;
}

ll comb(ll n, ll k, ll mod) {
    if (k > n || k < 0) return 0;
    if (k == n || k == 0) return 1;
    return fact[n] * invfact[n-k] % mod * invfact[k] % mod;
}

ll comb_rep(ll n, ll k, ll mod) {
    return comb(n+k-1, k, mod);
}


int main() {
    ll N, M;
    cin >> N >> M;
    ll mod = 998244353;
    set_fact(5000010, mod);
    
    ll ans = comb_rep(N, 3*M, mod);
    
    for (ll i = 2*M+1; i <= 3*M; i++) {
        ll ex = N*comb_rep(N-1, M*3-i, mod) % mod;
        ans += mod - ex;
    }
    
    for (ll i = M+1; i <= N; i++) {
        if ((3*M-i)%2 == 1) continue;
        if (i > M*3) break;
        ll ex = comb(N, i, mod) * comb_rep(N, (3*M-i)/2, mod) % mod;
        ans += mod - ex;
    }
    cout << ans%mod << endl;
    return 0;
}

### Prompt

In CPP, your task is to solve 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;
#define ll long long

ll modpower(ll a, ll r, ll mod){ //a^r
    ll x = 1;
    a %= mod;
    while (r > 0) {
        if (r & 1) x = x * a % mod;
        r >>= 1;
        a = a * a % mod;
    }
    return x;
}

ll modinv(ll a, ll mod){
    return modpower(a, mod-2, mod);
}

vector<ll> fact;
vector<ll> invfact;

void set_fact(ll n, ll mod){
    fact.resize(n+1, 1);
    invfact.resize(n+1, 1);
    for (ll i = 2; i <= n; i++) {
        fact[i] = fact[i-1] * i % mod;
    }
    invfact[n] = modinv(fact[n], mod);
    for (ll i = n-1; i >= 2; i--) {
        invfact[i] = invfact[i+1] * (i+1) % mod;
    }
    return;
}

ll comb(ll n, ll k, ll mod) {
    if (k > n || k < 0) return 0;
    if (k == n || k == 0) return 1;
    return fact[n] * invfact[n-k] % mod * invfact[k] % mod;
}

ll comb_rep(ll n, ll k, ll mod) {
    return comb(n+k-1, k, mod);
}


int main() {
    ll N, M;
    cin >> N >> M;
    ll mod = 998244353;
    set_fact(5000010, mod);
    
    ll ans = comb_rep(N, 3*M, mod);
    
    for (ll i = 2*M+1; i <= 3*M; i++) {
        ll ex = N*comb_rep(N-1, M*3-i, mod) % mod;
        ans += mod - ex;
    }
    
    for (ll i = M+1; i <= N; i++) {
        if ((3*M-i)%2 == 1) continue;
        if (i > M*3) break;
        ll ex = comb(N, i, mod) * comb_rep(N, (3*M-i)/2, mod) % mod;
        ans += mod - ex;
    }
    cout << ans%mod << endl;
    return 0;
}

```