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

const ll MAX = 3500100, MOD = 998244353;
ll fac[MAX], finv[MAX], inv[MAX];
// 前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD/i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
// 二項係数の計算
ll COM(int n, int k){
    if (n < 0 || k < 0 || n < k) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

int main(){
    int n, m; cin >> n >> m;
    COMinit();
    ll ans = 0;
    if(m%2 == 0){
        for(int i = 0; i <= m; i += 2){
            ans += COM((3*m-i)/2 + n-1, n-1) * COM(n, i) % MOD; // 奇数がi個
            ans -= COM((m-2-i)/2 + n-1, n-1) * COM(n, i) % MOD * n % MOD; // 2n+2以上
            ans -= COM((m-i)/2 + n-2, n-2) * COM(n-1, i-1) % MOD * n % MOD; // 2n+1
            ans += MOD;
            ans %= MOD;
        }
    }
    if(m%2 == 1){
        for(int i = 1; i <= m; i += 2){
            ans += COM((3*m-i)/2 + n-1, n-1) * COM(n, i) % MOD; // 奇数がi個
            ans -= COM((m-2-i)/2 + n-1, n-1) * COM(n, i) % MOD * n % MOD; // 2n+2以上
            ans -= COM((m-i)/2 + n-2, n-2) * COM(n-1, i-1) % MOD * n % MOD; // 2n+1
            ans += MOD;
            ans %= MOD;
        }
    }
    cout << ans << endl;
}
### Prompt

Please formulate 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 <bits/stdc++.h>
using ll = long long;
using namespace std;

const ll MAX = 3500100, MOD = 998244353;
ll fac[MAX], finv[MAX], inv[MAX];
// 前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD/i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
// 二項係数の計算
ll COM(int n, int k){
    if (n < 0 || k < 0 || n < k) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

int main(){
    int n, m; cin >> n >> m;
    COMinit();
    ll ans = 0;
    if(m%2 == 0){
        for(int i = 0; i <= m; i += 2){
            ans += COM((3*m-i)/2 + n-1, n-1) * COM(n, i) % MOD; // 奇数がi個
            ans -= COM((m-2-i)/2 + n-1, n-1) * COM(n, i) % MOD * n % MOD; // 2n+2以上
            ans -= COM((m-i)/2 + n-2, n-2) * COM(n-1, i-1) % MOD * n % MOD; // 2n+1
            ans += MOD;
            ans %= MOD;
        }
    }
    if(m%2 == 1){
        for(int i = 1; i <= m; i += 2){
            ans += COM((3*m-i)/2 + n-1, n-1) * COM(n, i) % MOD; // 奇数がi個
            ans -= COM((m-2-i)/2 + n-1, n-1) * COM(n, i) % MOD * n % MOD; // 2n+2以上
            ans -= COM((m-i)/2 + n-2, n-2) * COM(n-1, i-1) % MOD * n % MOD; // 2n+1
            ans += MOD;
            ans %= MOD;
        }
    }
    cout << ans << endl;
}
```