#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#define ll long long
#define mod 998244353
#define Mod1(x) (x>=mod?x-mod:x)
#define Mod2(x) (x<0?x+mod:x)
#define maxn 3000010
inline ll read()
{
	ll x=0; char c=getchar(),f=1;
	for(;c<'0'||'9'<c;c=getchar())if(c=='-')f=-1;
	for(;'0'<=c&&c<='9';c=getchar())x=x*10+c-'0';
	return x*f;
}
inline void write(ll x)
{
	static int buf[20],len; len=0;
	if(x<0)x=-x,putchar('-');
	for(;x;x/=10)buf[len++]=x%10;
	if(!len)putchar('0');
	else while(len)putchar(buf[--len]+'0');
}
inline void writeln(ll x){write(x); putchar('\n');}
inline void writesp(ll x){write(x); putchar(' ');}
ll fac[maxn],inv[maxn];
int n,m;
inline ll power(ll a,ll b)
{
	ll ans=1;
	for(;b;b>>=1,a=a*a%mod)
		if(b&1)ans=ans*a%mod;
	return ans;
}
inline ll C(int n,int m){return fac[n]*inv[m]%mod*inv[n-m]%mod;}
inline ll calc(int n,int m,int k)
{
	ll sum=0;
	for(int i=0;i<=n&&i<=k;i++)
		if(!((m-i)&1)&&m>=i)sum=(sum+C(n,i)*C((m-i)/2+n-1,n-1))%mod;
	// writeln(sum);
	return sum;
}
int main()
{
	n=read(); m=read();
	fac[0]=inv[0]=1;
	for(int i=1;i<=n+3*m;i++){
		fac[i]=fac[i-1]*i%mod;
		inv[i]=power(fac[i],mod-2);
	}
	ll ans=(calc(n,3*m,m)-n*calc(n,m-1,m))%mod;
	writeln(Mod2(ans));
	return 0;
}
### Prompt

Please formulate 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<cmath>
#include<cstdlib>
#include<algorithm>
#define ll long long
#define mod 998244353
#define Mod1(x) (x>=mod?x-mod:x)
#define Mod2(x) (x<0?x+mod:x)
#define maxn 3000010
inline ll read()
{
	ll x=0; char c=getchar(),f=1;
	for(;c<'0'||'9'<c;c=getchar())if(c=='-')f=-1;
	for(;'0'<=c&&c<='9';c=getchar())x=x*10+c-'0';
	return x*f;
}
inline void write(ll x)
{
	static int buf[20],len; len=0;
	if(x<0)x=-x,putchar('-');
	for(;x;x/=10)buf[len++]=x%10;
	if(!len)putchar('0');
	else while(len)putchar(buf[--len]+'0');
}
inline void writeln(ll x){write(x); putchar('\n');}
inline void writesp(ll x){write(x); putchar(' ');}
ll fac[maxn],inv[maxn];
int n,m;
inline ll power(ll a,ll b)
{
	ll ans=1;
	for(;b;b>>=1,a=a*a%mod)
		if(b&1)ans=ans*a%mod;
	return ans;
}
inline ll C(int n,int m){return fac[n]*inv[m]%mod*inv[n-m]%mod;}
inline ll calc(int n,int m,int k)
{
	ll sum=0;
	for(int i=0;i<=n&&i<=k;i++)
		if(!((m-i)&1)&&m>=i)sum=(sum+C(n,i)*C((m-i)/2+n-1,n-1))%mod;
	// writeln(sum);
	return sum;
}
int main()
{
	n=read(); m=read();
	fac[0]=inv[0]=1;
	for(int i=1;i<=n+3*m;i++){
		fac[i]=fac[i-1]*i%mod;
		inv[i]=power(fac[i],mod-2);
	}
	ll ans=(calc(n,3*m,m)-n*calc(n,m-1,m))%mod;
	writeln(Mod2(ans));
	return 0;
}
```