#include <bits/stdc++.h>
#define sz(c)      int(c.size())
#define rep(i,a,b) for (int i=a; i<(b); ++i)
#define per(i,a,b) for (int i=(b)-1; i>=(a); --i)
using namespace std;
using ll = long long;

int const MAXN=2.2e6;
int const MD=998244353;

int madd(int x,int y) {
	if (x+y<MD) return x+y; else return x+y-MD;
}
int msub(int x,int y) {
	if (x-y>=0) return x-y; else return x-y+MD;
}
int mmul(int x,int y) {
	return int(ll(x)*y % MD);
}
int mpow(int x,int p) {
	int res=1;
	while (p>0) {
		if (p%2) res=mmul(res,x);
		x=mmul(x,x);
		p/=2;
	}
	return res;
}
int minv(int x) {
	return mpow(x,MD-2);
}
int mdiv(int x,int y) {
	return mmul(x,mpow(y,MD-2));
}

struct mod_fact {
	vector<int> f,rf;
	mod_fact(int n):f(n),rf(n) {
		f[0]=rf[0]=1;
		rep(i,1,n) {
			f[i]=mmul(f[i-1],i);
			rf[i]=minv(f[i]);
		}
	}
} mfact(MAXN);

int comb(int n,int k) {
	if (n<k) return 0;
	return mmul(mfact.f[n],mmul(mfact.rf[n-k],mfact.rf[k]));
}
int distr(int n,int k) {
	return comb(n+k-1,k-1);
}

int main() {
	cin.tie(0);
	ios_base::sync_with_stdio(0);
	cout<<fixed<<setprecision(10);

	int N,M;
	cin>>N>>M;
	
	int res=0;
	rep(i,0,min(N,M)+1) if ((3*M-i)%2==0) {
		int add=mmul(comb(N,i),distr((3*M-i)/2,N));
		res=madd(res,add);
	}
	rep(i,0,min(N,M)+1) if ((M-i)%2==0) {
		int sub=mmul(comb(N,i),distr((M-i)/2,N));
		res=msub(res,mmul(sub,N));
	}
	rep(i,0,min(N,M)+1) if ((M-i)%2==0) {
		int add=mmul(comb(N-1,i),distr((M-i)/2,N-1));
		res=madd(res,mmul(add,N));
	}
	cout<<res<<'\n';
}

### Prompt

Create a solution in Cpp for 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>
#define sz(c)      int(c.size())
#define rep(i,a,b) for (int i=a; i<(b); ++i)
#define per(i,a,b) for (int i=(b)-1; i>=(a); --i)
using namespace std;
using ll = long long;

int const MAXN=2.2e6;
int const MD=998244353;

int madd(int x,int y) {
	if (x+y<MD) return x+y; else return x+y-MD;
}
int msub(int x,int y) {
	if (x-y>=0) return x-y; else return x-y+MD;
}
int mmul(int x,int y) {
	return int(ll(x)*y % MD);
}
int mpow(int x,int p) {
	int res=1;
	while (p>0) {
		if (p%2) res=mmul(res,x);
		x=mmul(x,x);
		p/=2;
	}
	return res;
}
int minv(int x) {
	return mpow(x,MD-2);
}
int mdiv(int x,int y) {
	return mmul(x,mpow(y,MD-2));
}

struct mod_fact {
	vector<int> f,rf;
	mod_fact(int n):f(n),rf(n) {
		f[0]=rf[0]=1;
		rep(i,1,n) {
			f[i]=mmul(f[i-1],i);
			rf[i]=minv(f[i]);
		}
	}
} mfact(MAXN);

int comb(int n,int k) {
	if (n<k) return 0;
	return mmul(mfact.f[n],mmul(mfact.rf[n-k],mfact.rf[k]));
}
int distr(int n,int k) {
	return comb(n+k-1,k-1);
}

int main() {
	cin.tie(0);
	ios_base::sync_with_stdio(0);
	cout<<fixed<<setprecision(10);

	int N,M;
	cin>>N>>M;
	
	int res=0;
	rep(i,0,min(N,M)+1) if ((3*M-i)%2==0) {
		int add=mmul(comb(N,i),distr((3*M-i)/2,N));
		res=madd(res,add);
	}
	rep(i,0,min(N,M)+1) if ((M-i)%2==0) {
		int sub=mmul(comb(N,i),distr((M-i)/2,N));
		res=msub(res,mmul(sub,N));
	}
	rep(i,0,min(N,M)+1) if ((M-i)%2==0) {
		int add=mmul(comb(N-1,i),distr((M-i)/2,N-1));
		res=madd(res,mmul(add,N));
	}
	cout<<res<<'\n';
}

```