#include <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <cmath>
#include <bitset>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <algorithm>
#include <complex>
#include <unordered_map>
#include <unordered_set>
#include <random>
#include <cassert>
#include <fstream>
#define popcount __builtin_popcount
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
const ll MOD=998244353;
ll powmod(ll a, ll k){
    ll ap=a, ans=1;
    while(k){
        if(k&1){
            ans*=ap;
            ans%=MOD;
        }
        ap=ap*ap;
        ap%=MOD;
        k>>=1;
    }
    return ans;
}
ll inv(ll a){
	return powmod(a, MOD-2);
}
ll f[5000001], invf[5000001];
void fac(int n){
    f[0]=1;
    for(ll i=1; i<=n; i++) f[i]=f[i-1]*i%MOD;
    invf[n]=inv(f[n]);
    for(ll i=n-1; i>=0; i--) invf[i]=invf[i+1]*(i+1)%MOD;
}
ll comb(int x, int y){
    if(!(0<=y && y<=x)) return 0;
    return f[x]*invf[y]%MOD*invf[x-y]%MOD;
}
int main()
{
    int n, m; cin>>n>>m;
    fac(3*m+n);
    ll ans=0;
    for(int i=0; i<=m; i++){
        if((m-i)%2!=0) continue;
        int l=(3*m-i)/2;
        (ans+=comb(n, i)*comb(l+n-1, n-1))%=MOD;
        (ans+=MOD-i*comb(n, i)%MOD*comb(l-m+n-1, n-1)%MOD)%=MOD;
        (ans+=MOD-(n-i)*comb(n, i)%MOD*comb(l-m-1+n-1, n-1)%MOD)%=MOD;
    }
    cout<<ans<<endl;
    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 <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <cmath>
#include <bitset>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <algorithm>
#include <complex>
#include <unordered_map>
#include <unordered_set>
#include <random>
#include <cassert>
#include <fstream>
#define popcount __builtin_popcount
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
const ll MOD=998244353;
ll powmod(ll a, ll k){
    ll ap=a, ans=1;
    while(k){
        if(k&1){
            ans*=ap;
            ans%=MOD;
        }
        ap=ap*ap;
        ap%=MOD;
        k>>=1;
    }
    return ans;
}
ll inv(ll a){
	return powmod(a, MOD-2);
}
ll f[5000001], invf[5000001];
void fac(int n){
    f[0]=1;
    for(ll i=1; i<=n; i++) f[i]=f[i-1]*i%MOD;
    invf[n]=inv(f[n]);
    for(ll i=n-1; i>=0; i--) invf[i]=invf[i+1]*(i+1)%MOD;
}
ll comb(int x, int y){
    if(!(0<=y && y<=x)) return 0;
    return f[x]*invf[y]%MOD*invf[x-y]%MOD;
}
int main()
{
    int n, m; cin>>n>>m;
    fac(3*m+n);
    ll ans=0;
    for(int i=0; i<=m; i++){
        if((m-i)%2!=0) continue;
        int l=(3*m-i)/2;
        (ans+=comb(n, i)*comb(l+n-1, n-1))%=MOD;
        (ans+=MOD-i*comb(n, i)%MOD*comb(l-m+n-1, n-1)%MOD)%=MOD;
        (ans+=MOD-(n-i)*comb(n, i)%MOD*comb(l-m-1+n-1, n-1)%MOD)%=MOD;
    }
    cout<<ans<<endl;
    return 0;
}
```