#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define N 4000400
const int mod=998244353;
typedef long long ll;
inline ll read(){
	ll x=0,f=1;
	char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-')f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9'){
		x=(x<<1)+(x<<3)+c-'0';
		c=getchar();
	}
	return x*f;
}
int n,m,ifac[N],fac[N],ans;
int qpow(int a,int b){
	int ans=1;
	while(b){
		if(b&1)ans=1LL*ans*a%mod;
		a=1LL*a*a%mod;
		b>>=1;
	}
	return ans;
}
void init(int n){
	fac[0]=1;
	for(int i=1;i<=n;++i){
		fac[i]=1LL*fac[i-1]*i%mod;
	}
	ifac[n]=qpow(fac[n],mod-2)%mod;
	for(int i=n-1;i>=0;--i){
		ifac[i]=1LL*ifac[i+1]*(i+1)%mod;
	}
}
inline int C(int n,int m){
	if(n<0||m<0||n-m<0)return 0;
	return 1LL*fac[n]*ifac[m]%mod*ifac[n-m]%mod;
}
inline int calc(int n,int m){
	return C(n+m-1,n-1);
}
int main(){
	n=read(),m=read();
	init(n+3*m);
	for(int i=0;i<=min(n,m);++i){
		if((i&1)^(m&1))continue;
		ans=(ans+1LL*C(n,i)*calc(n,(3*m-i)>>1))%mod;
		ans=(ans-1LL*C(n-1,i)*n%mod*calc(n,(m-i-2)>>1))%mod;
		ans=(ans-1LL*C(n-1,i-1)*n%mod*calc(n,(m-i)>>1))%mod;
	}
	printf("%d\n",(ans+mod)%mod);
	return 0;
}



### Prompt

Please provide a CPP coded solution to the problem described below:

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>
using namespace std;
#define N 4000400
const int mod=998244353;
typedef long long ll;
inline ll read(){
	ll x=0,f=1;
	char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-')f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9'){
		x=(x<<1)+(x<<3)+c-'0';
		c=getchar();
	}
	return x*f;
}
int n,m,ifac[N],fac[N],ans;
int qpow(int a,int b){
	int ans=1;
	while(b){
		if(b&1)ans=1LL*ans*a%mod;
		a=1LL*a*a%mod;
		b>>=1;
	}
	return ans;
}
void init(int n){
	fac[0]=1;
	for(int i=1;i<=n;++i){
		fac[i]=1LL*fac[i-1]*i%mod;
	}
	ifac[n]=qpow(fac[n],mod-2)%mod;
	for(int i=n-1;i>=0;--i){
		ifac[i]=1LL*ifac[i+1]*(i+1)%mod;
	}
}
inline int C(int n,int m){
	if(n<0||m<0||n-m<0)return 0;
	return 1LL*fac[n]*ifac[m]%mod*ifac[n-m]%mod;
}
inline int calc(int n,int m){
	return C(n+m-1,n-1);
}
int main(){
	n=read(),m=read();
	init(n+3*m);
	for(int i=0;i<=min(n,m);++i){
		if((i&1)^(m&1))continue;
		ans=(ans+1LL*C(n,i)*calc(n,(3*m-i)>>1))%mod;
		ans=(ans-1LL*C(n-1,i)*n%mod*calc(n,(m-i-2)>>1))%mod;
		ans=(ans-1LL*C(n-1,i-1)*n%mod*calc(n,(m-i)>>1))%mod;
	}
	printf("%d\n",(ans+mod)%mod);
	return 0;
}



```