#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
typedef long long ll;
ll fac[2000005], inv[2000005];
ll exp(ll a, int x){
    if (x == 0) return 1ll;
    ll p = exp(a,x/2);
    p = (p*p)%mod;
    if (x&1) p = (p*a)%mod;
    return p;
}
ll C(ll n, ll k){
    return ((fac[n]*inv[n-k])%mod * inv[k])%mod;
}
ll f(int n, int s, int k){
    //printf("f(%d,%d,%d):\n",n,s,k);
    ll ret= 0;
    for (int i = 0; i <= min(k,n); i++){
        int _s = s-i;
        if (_s % 2 != 0) continue;
        //printf("can take %d odd\n",i);
        ret += (C(n,i)*C(_s/2+n-1,n-1))%mod;
    }
    ret %= mod;
    //printf("%d elements %d sum %d odd -> %lld\n",n,s,k,ret);
    return ret;
}
int n,m;
int main(){
    ///X : sum = 3M
    ///Y : none > 2M
    ///Z : <= M odd
    scanf("%d%d",&n,&m);
    fac[0] = inv[0] = 1;
    for (int i = 1; i <= 2000000; i++){
        fac[i] = (fac[i-1]*i)%mod;
        inv[i] = exp(fac[i],mod-2);
    }
    ll XZ = f(n,3*m,m);
    ll XyZ1 = f(n,m,m);
    ll XyZ2 = f(n-1,m,m);
    ll XyZ = (n*(XyZ1 - XyZ2))%mod;
    ll ans = XZ - XyZ;
    ans = ((ans%mod)+mod)%mod;
    //printf("%lld %lld %lld\n",XZ,XyZ1,XyZ2);
    printf("%lld",ans);

}

### 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;
const int mod = 998244353;
typedef long long ll;
ll fac[2000005], inv[2000005];
ll exp(ll a, int x){
    if (x == 0) return 1ll;
    ll p = exp(a,x/2);
    p = (p*p)%mod;
    if (x&1) p = (p*a)%mod;
    return p;
}
ll C(ll n, ll k){
    return ((fac[n]*inv[n-k])%mod * inv[k])%mod;
}
ll f(int n, int s, int k){
    //printf("f(%d,%d,%d):\n",n,s,k);
    ll ret= 0;
    for (int i = 0; i <= min(k,n); i++){
        int _s = s-i;
        if (_s % 2 != 0) continue;
        //printf("can take %d odd\n",i);
        ret += (C(n,i)*C(_s/2+n-1,n-1))%mod;
    }
    ret %= mod;
    //printf("%d elements %d sum %d odd -> %lld\n",n,s,k,ret);
    return ret;
}
int n,m;
int main(){
    ///X : sum = 3M
    ///Y : none > 2M
    ///Z : <= M odd
    scanf("%d%d",&n,&m);
    fac[0] = inv[0] = 1;
    for (int i = 1; i <= 2000000; i++){
        fac[i] = (fac[i-1]*i)%mod;
        inv[i] = exp(fac[i],mod-2);
    }
    ll XZ = f(n,3*m,m);
    ll XyZ1 = f(n,m,m);
    ll XyZ2 = f(n-1,m,m);
    ll XyZ = (n*(XyZ1 - XyZ2))%mod;
    ll ans = XZ - XyZ;
    ans = ((ans%mod)+mod)%mod;
    //printf("%lld %lld %lld\n",XZ,XyZ1,XyZ2);
    printf("%lld",ans);

}

```