#include<bits/stdc++.h>
using namespace std;
#define mod 998244353
#define MN 2500000
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
int n,m,ans,p[MN+5],inv[MN+5];
inline int C(int n,int m){return m>n?0:1LL*p[n]*inv[m]%mod*inv[n-m]%mod;}
int main()
{
    n=read();m=read();
    p[0]=p[1]=inv[0]=inv[1]=1;
    for(int i=2;i<=MN;++i) p[i]=1LL*p[i-1]*i%mod,inv[i]=1LL*(mod-mod/i)*inv[mod%i]%mod;
    for(int i=2;i<=MN;++i) inv[i]=1LL*inv[i-1]*inv[i]%mod;
    if(n==2) return 0*printf("%d\n",m+1);
    for(int i=0;i<=m;i+=2) if(m-i<=n)
    {
        int t=i/2+m;
        ans=(ans+1LL*C(n,m-i)*C(t+n-1,n-1))%mod;
        ans=(ans+1LL*C(n,m-i)*(mod-(m-i))%mod*C(t-m+n-1,n-1))%mod;
        ans=(ans+1LL*C(n,m-i)*(mod-(n-(m-i)))%mod*C(t-m+n-2,n-1))%mod;
    }
    cout<<ans;
    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<bits/stdc++.h>
using namespace std;
#define mod 998244353
#define MN 2500000
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
int n,m,ans,p[MN+5],inv[MN+5];
inline int C(int n,int m){return m>n?0:1LL*p[n]*inv[m]%mod*inv[n-m]%mod;}
int main()
{
    n=read();m=read();
    p[0]=p[1]=inv[0]=inv[1]=1;
    for(int i=2;i<=MN;++i) p[i]=1LL*p[i-1]*i%mod,inv[i]=1LL*(mod-mod/i)*inv[mod%i]%mod;
    for(int i=2;i<=MN;++i) inv[i]=1LL*inv[i-1]*inv[i]%mod;
    if(n==2) return 0*printf("%d\n",m+1);
    for(int i=0;i<=m;i+=2) if(m-i<=n)
    {
        int t=i/2+m;
        ans=(ans+1LL*C(n,m-i)*C(t+n-1,n-1))%mod;
        ans=(ans+1LL*C(n,m-i)*(mod-(m-i))%mod*C(t-m+n-1,n-1))%mod;
        ans=(ans+1LL*C(n,m-i)*(mod-(n-(m-i)))%mod*C(t-m+n-2,n-1))%mod;
    }
    cout<<ans;
    return 0;
}
```