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

#define rep(i,n) for(int i=0; i<(n); ++i)
#define rrep(i,n) for(int i=1; i<=(n); ++i)
#define drep(i,n) for(int i=(n)-1; i>=0; --i)
#define foreach(it, v) for(typeof(v.begin()) it=v.begin(); it!=v.end(); ++it)
#define pb push_back
#define sz(x) (int)(x).size()
#define fi first
#define se second

typedef unsigned int uint;
typedef long long int ll;
typedef vector<int> vi;
typedef pair<int,int> pii;
typedef vector<pii> vpii;
typedef set<int> si;


const int MAX = 3010000;
const int MOD = 998244353;

ll fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
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;
    }
}

// 二項係数計算
ll COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}






int N, M;

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);

  // 前処理
  COMinit();

  cin >> N >> M;

  ll ans = 0;
  for (int n=M%2; n<=M; n += 2) {
    int D = (M * 3 - n) / 2;
    ll a1 = COM(N, n) * COM(N+D-1, N-1) % MOD;
    ll b1 = N * (COM(N-1, n-1) * COM(N+D-M-1, N-1) % MOD) % MOD;
    ll b2 = N * (COM(N-1, n) * COM(N+D-M-2, N-1) % MOD) % MOD;
    ans = (ans + a1 - b1 - b2 + MOD*2) % MOD;
  }
  cout << ans << "\n";
  return 0;
}

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

#define rep(i,n) for(int i=0; i<(n); ++i)
#define rrep(i,n) for(int i=1; i<=(n); ++i)
#define drep(i,n) for(int i=(n)-1; i>=0; --i)
#define foreach(it, v) for(typeof(v.begin()) it=v.begin(); it!=v.end(); ++it)
#define pb push_back
#define sz(x) (int)(x).size()
#define fi first
#define se second

typedef unsigned int uint;
typedef long long int ll;
typedef vector<int> vi;
typedef pair<int,int> pii;
typedef vector<pii> vpii;
typedef set<int> si;


const int MAX = 3010000;
const int MOD = 998244353;

ll fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
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;
    }
}

// 二項係数計算
ll COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}






int N, M;

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);

  // 前処理
  COMinit();

  cin >> N >> M;

  ll ans = 0;
  for (int n=M%2; n<=M; n += 2) {
    int D = (M * 3 - n) / 2;
    ll a1 = COM(N, n) * COM(N+D-1, N-1) % MOD;
    ll b1 = N * (COM(N-1, n-1) * COM(N+D-M-1, N-1) % MOD) % MOD;
    ll b2 = N * (COM(N-1, n) * COM(N+D-M-2, N-1) % MOD) % MOD;
    ans = (ans + a1 - b1 - b2 + MOD*2) % MOD;
  }
  cout << ans << "\n";
  return 0;
}

```