#include <iostream>
#include <cstdio>
#include <cstring>
#include <string.h>
#include <cstdlib>
#define N 2000005
using namespace std;
typedef long long ll;
const int mod=998244353;
int n,m;
ll f[N],inv[N];
ll C(int n,int m){
	return f[n]*inv[m]%mod*inv[n-m]%mod;
}
ll solve(int n,int m){
	ll ans=0;
	for(int i=m&1;i<=m&&i<=n;i+=2){
		ll w=C(n,i);
		int res=m*3-i>>1;
		ans=(ans+w*C(res+n-1,n-1))%mod;
	}
	return ans;
}
int main(){
	scanf("%d%d",&n,&m);
	f[0]=f[1]=inv[0]=inv[1]=1;
	for(int i=2;i<=2000002;i++) f[i]=f[i-1]*i%mod,inv[i]=(mod-mod/i)*inv[mod%i]%mod;
	for(int i=1;i<=2000002;i++) inv[i]=inv[i-1]*inv[i]%mod;
	ll ans=(solve(n,m)-C(n+m-2,n-1)*n)%mod;
	ans=(ans+mod)%mod;
	printf("%lld",ans);
	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 <iostream>
#include <cstdio>
#include <cstring>
#include <string.h>
#include <cstdlib>
#define N 2000005
using namespace std;
typedef long long ll;
const int mod=998244353;
int n,m;
ll f[N],inv[N];
ll C(int n,int m){
	return f[n]*inv[m]%mod*inv[n-m]%mod;
}
ll solve(int n,int m){
	ll ans=0;
	for(int i=m&1;i<=m&&i<=n;i+=2){
		ll w=C(n,i);
		int res=m*3-i>>1;
		ans=(ans+w*C(res+n-1,n-1))%mod;
	}
	return ans;
}
int main(){
	scanf("%d%d",&n,&m);
	f[0]=f[1]=inv[0]=inv[1]=1;
	for(int i=2;i<=2000002;i++) f[i]=f[i-1]*i%mod,inv[i]=(mod-mod/i)*inv[mod%i]%mod;
	for(int i=1;i<=2000002;i++) inv[i]=inv[i-1]*inv[i]%mod;
	ll ans=(solve(n,m)-C(n+m-2,n-1)*n)%mod;
	ans=(ans+mod)%mod;
	printf("%lld",ans);
	return 0;
}

```