#include<bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i,n) for(int i=0; i<(n); i++)
const int MOD = 998244353, MX = 2e6+5;
int n,m;
ll exp(ll b, ll n){
	ll res = 1;
	for(; n; n/=2,(b*=b)%=MOD) if(n%2) (res *= b) %= MOD;
	return res;
}
ll fact[MX] = {1}, invf[MX];
void setcomb(int n){
	for(int i = 1; i <= n; i++) fact[i] = (fact[i-1] * i) % MOD;
	invf[n] =  exp(fact[n] ,MOD-2);
	for(int i = n; i > 0; i--) invf[i-1] = (invf[i] * i) % MOD;
}
inline ll comb(int a, int b){return fact[a] * invf[a-b] % MOD * invf[b] % MOD;}
ll ans;
int main(){
	scanf("%d%d", &n, &m);
	setcomb(m*2+n);
	ans = MOD - n*comb(m+n-2,n-1)%MOD;
	for(int i=m&1; i<=min(m,n); i+=2) ans += comb(n,i) * comb((m*3-i)/2+n-1,n-1) % MOD;
	printf("%lld\n", ans % MOD);
}
### Prompt

Please create a solution in Cpp 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;
using ll = long long;
#define rep(i,n) for(int i=0; i<(n); i++)
const int MOD = 998244353, MX = 2e6+5;
int n,m;
ll exp(ll b, ll n){
	ll res = 1;
	for(; n; n/=2,(b*=b)%=MOD) if(n%2) (res *= b) %= MOD;
	return res;
}
ll fact[MX] = {1}, invf[MX];
void setcomb(int n){
	for(int i = 1; i <= n; i++) fact[i] = (fact[i-1] * i) % MOD;
	invf[n] =  exp(fact[n] ,MOD-2);
	for(int i = n; i > 0; i--) invf[i-1] = (invf[i] * i) % MOD;
}
inline ll comb(int a, int b){return fact[a] * invf[a-b] % MOD * invf[b] % MOD;}
ll ans;
int main(){
	scanf("%d%d", &n, &m);
	setcomb(m*2+n);
	ans = MOD - n*comb(m+n-2,n-1)%MOD;
	for(int i=m&1; i<=min(m,n); i+=2) ans += comb(n,i) * comb((m*3-i)/2+n-1,n-1) % MOD;
	printf("%lld\n", ans % MOD);
}
```