#include<bits/stdc++.h>
using namespace std;
const int N=5e6+10;
const int mod=998244353;

int gi() {
	int x=0,o=1;char ch=getchar();
	while(!isdigit(ch)&&ch!='-') ch=getchar();
	if(ch=='-') o=-1,ch=getchar();
	while(isdigit(ch)) x=x*10+ch-'0',ch=getchar();
	return x*o;
}

int n,m,fac[N],inv[N],ifac[N];

int C(int n,int m) {
	if(m<0||n<m) return 0;
	return 1ll*fac[n]*ifac[m]%mod*ifac[n-m]%mod;
}

int main() {
	cin>>n>>m;
	fac[0]=1;for(int i=1;i<N;i++) fac[i]=1ll*fac[i-1]*i%mod;
	inv[1]=1;for(int i=2;i<N;i++) inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
	ifac[0]=1;for(int i=1;i<N;i++) ifac[i]=1ll*ifac[i-1]*inv[i]%mod;
	int ans=0;
	for(int i=0;i<=m;i++) if(!((3*m-i)&1)) ans=(ans+1ll*C(n,i)*C((3*m-i)/2+n-1,n-1))%mod;
	for(int i=2*m+1;i<=3*m;i++) ans=(ans-1ll*n*C(3*m-i+n-2,n-2))%mod;
	cout<<(ans%mod+mod)%mod;
	return 0;
}

### Prompt

Generate 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;
const int N=5e6+10;
const int mod=998244353;

int gi() {
	int x=0,o=1;char ch=getchar();
	while(!isdigit(ch)&&ch!='-') ch=getchar();
	if(ch=='-') o=-1,ch=getchar();
	while(isdigit(ch)) x=x*10+ch-'0',ch=getchar();
	return x*o;
}

int n,m,fac[N],inv[N],ifac[N];

int C(int n,int m) {
	if(m<0||n<m) return 0;
	return 1ll*fac[n]*ifac[m]%mod*ifac[n-m]%mod;
}

int main() {
	cin>>n>>m;
	fac[0]=1;for(int i=1;i<N;i++) fac[i]=1ll*fac[i-1]*i%mod;
	inv[1]=1;for(int i=2;i<N;i++) inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
	ifac[0]=1;for(int i=1;i<N;i++) ifac[i]=1ll*ifac[i-1]*inv[i]%mod;
	int ans=0;
	for(int i=0;i<=m;i++) if(!((3*m-i)&1)) ans=(ans+1ll*C(n,i)*C((3*m-i)/2+n-1,n-1))%mod;
	for(int i=2*m+1;i<=3*m;i++) ans=(ans-1ll*n*C(3*m-i+n-2,n-2))%mod;
	cout<<(ans%mod+mod)%mod;
	return 0;
}

```