#include<cstdio>
const int MAX = 3000000;
const int MOD = 998244353;
int add(int a, int b) {return (a + b)%MOD;}
int mul(int a, int b) {return 1LL*a*b%MOD;}
int sub(int a, int b) {return add(a, MOD-b);}
int pow_mod(int b, int p) {
	int ret = 1;
	while( p ) {
		if( p & 1 ) ret = mul(ret, b);
		b = mul(b, b);
		p >>= 1;
	}
	return ret;
}
int fct[MAX + 5], ifct[MAX + 5];
void init() {
	fct[0] = 1;
	for(int i=1;i<=MAX;i++)
		fct[i] = mul(fct[i-1], i);
	ifct[MAX] = pow_mod(fct[MAX], MOD - 2);
	for(int i=MAX-1;i>=0;i--)
		ifct[i] = mul(ifct[i+1], i+1);
}
int comb(int n, int m) {
	if( n < m ) return 0;
	else return mul(fct[n], mul(ifct[m], ifct[n-m]));
}
int n, m, l;
int main() {
	init(); scanf("%d%d", &n, &m);
	int ans = 0;
	for(int i=(m&1);i<=n&&i<=m;i+=2) {
		int t = (3*m - i) / 2;
		int del = comb(t + n - 1, n - 1);
		if( t >= m )
			del = sub(del, mul(i, comb(t - m + n - 1, n - 1)));
		if( t >= m + 1 )
			del = sub(del, mul(n - i, comb(t - m - 1 + n - 1, n - 1)));
		ans = add(ans, mul(comb(n, i), del));
	}
	printf("%d\n", ans);
}
### Prompt

Your challenge is to write 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>
const int MAX = 3000000;
const int MOD = 998244353;
int add(int a, int b) {return (a + b)%MOD;}
int mul(int a, int b) {return 1LL*a*b%MOD;}
int sub(int a, int b) {return add(a, MOD-b);}
int pow_mod(int b, int p) {
	int ret = 1;
	while( p ) {
		if( p & 1 ) ret = mul(ret, b);
		b = mul(b, b);
		p >>= 1;
	}
	return ret;
}
int fct[MAX + 5], ifct[MAX + 5];
void init() {
	fct[0] = 1;
	for(int i=1;i<=MAX;i++)
		fct[i] = mul(fct[i-1], i);
	ifct[MAX] = pow_mod(fct[MAX], MOD - 2);
	for(int i=MAX-1;i>=0;i--)
		ifct[i] = mul(ifct[i+1], i+1);
}
int comb(int n, int m) {
	if( n < m ) return 0;
	else return mul(fct[n], mul(ifct[m], ifct[n-m]));
}
int n, m, l;
int main() {
	init(); scanf("%d%d", &n, &m);
	int ans = 0;
	for(int i=(m&1);i<=n&&i<=m;i+=2) {
		int t = (3*m - i) / 2;
		int del = comb(t + n - 1, n - 1);
		if( t >= m )
			del = sub(del, mul(i, comb(t - m + n - 1, n - 1)));
		if( t >= m + 1 )
			del = sub(del, mul(n - i, comb(t - m - 1 + n - 1, n - 1)));
		ans = add(ans, mul(comb(n, i), del));
	}
	printf("%d\n", ans);
}
```