#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll mod=998244353;
const int MAXN=3e6+6;
ll ans,inv[MAXN],fac[MAXN],rfac[MAXN];
ll n,m,lim;
ll C(ll n,ll m){
	if (n<m) return 0;
	return fac[n]*rfac[m]%mod*rfac[n-m]%mod;
}
ll calc(ll n,ll M,ll m){
	ll ret=0;
	for (ll i=0;i<=m;i++){
		if ((M-i)%2==0){
			ret=(ret+C(n,i)*C((M-i)/2+n-1,n-1)%mod)%mod;
		}
	}
	return ret;
}
int main(){
	scanf("%lld%lld",&n,&m);
	lim=n+3*m; inv[1]=1;
	for (int i=2;i<=lim;i++){
		inv[i]=(mod-(mod/i)*inv[mod%i])%mod;
	}
	fac[0]=rfac[0]=1;
	for (int i=1;i<=lim;i++){
		fac[i]=fac[i-1]*i%mod;
		rfac[i]=rfac[i-1]*inv[i]%mod;
	}
	ans=(calc(n,3*m,m)-n*(calc(n,m,m)-calc(n-1,m,m))%mod)%mod;
	printf("%lld\n",(ans+mod)%mod);
	return 0;
}

### Prompt

Create a solution in cpp for 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<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll mod=998244353;
const int MAXN=3e6+6;
ll ans,inv[MAXN],fac[MAXN],rfac[MAXN];
ll n,m,lim;
ll C(ll n,ll m){
	if (n<m) return 0;
	return fac[n]*rfac[m]%mod*rfac[n-m]%mod;
}
ll calc(ll n,ll M,ll m){
	ll ret=0;
	for (ll i=0;i<=m;i++){
		if ((M-i)%2==0){
			ret=(ret+C(n,i)*C((M-i)/2+n-1,n-1)%mod)%mod;
		}
	}
	return ret;
}
int main(){
	scanf("%lld%lld",&n,&m);
	lim=n+3*m; inv[1]=1;
	for (int i=2;i<=lim;i++){
		inv[i]=(mod-(mod/i)*inv[mod%i])%mod;
	}
	fac[0]=rfac[0]=1;
	for (int i=1;i<=lim;i++){
		fac[i]=fac[i-1]*i%mod;
		rfac[i]=rfac[i-1]*inv[i]%mod;
	}
	ans=(calc(n,3*m,m)-n*(calc(n,m,m)-calc(n-1,m,m))%mod)%mod;
	printf("%lld\n",(ans+mod)%mod);
	return 0;
}

```