#include <bits/stdc++.h>
#define LL long long
#define P pair<int, int>
const LL N = 4e6 + 10;
const LL inf = 0x3f3f3f;
const LL mod = 998244353;
using namespace std;
template <typename tp>
inline void read(tp &x)
{
    x = 0; char c = getchar(); bool f = 0;
    for (; c < '0' || c > '9'; f |= (c == '-'), c = getchar()) ;
    for (; c >= '0' && c <= '9'; x = (x << 3) + (x << 1) + c - '0', c = getchar()) ;
    if (f) x = -x;
}
LL Ksm(LL a,LL b,LL ret=1)
{
    while(b)
    {
        if(b&1) ret=a*ret%mod;
        a=a*a%mod;
        b>>=1;
    }
    return ret;
}
LL n,m,ans;
LL inv[N],jc[N];
void init()
{
    inv[0]=jc[0]=jc[1]=1;
    for(int i=2;i<=4000000;i++) jc[i]=jc[i-1]*i%mod;
    inv[4000000]=Ksm(jc[4000000],mod-2);
    for(int i=3999999;i>=1;i--) inv[i]=inv[i+1]*(i+1)%mod;
}
LL C(LL x,LL y){return jc[y]*inv[y-x]%mod*inv[x]%mod;}
int main()
{
    read(n),read(m);
    init();
    for(int i=0;i<=m/2;i++)
    {
        LL ji=i*2+m%2,ou=m+(m-ji)/2;
        if(ji>n) break;
        ans+=C(ji,n)*C(n-1,ou+n-1)%mod;
        ans%=mod;
    }
    for(int i=m*2+1;i<=3*m;i++)
    {
        ans-=n*C(n-2,3*m-i+n-2)%mod;
        ans=(ans+mod)%mod;
    }
    printf("%lld\n",ans);
    return 0;
}
### Prompt

In cpp, your task is to solve 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 LL long long
#define P pair<int, int>
const LL N = 4e6 + 10;
const LL inf = 0x3f3f3f;
const LL mod = 998244353;
using namespace std;
template <typename tp>
inline void read(tp &x)
{
    x = 0; char c = getchar(); bool f = 0;
    for (; c < '0' || c > '9'; f |= (c == '-'), c = getchar()) ;
    for (; c >= '0' && c <= '9'; x = (x << 3) + (x << 1) + c - '0', c = getchar()) ;
    if (f) x = -x;
}
LL Ksm(LL a,LL b,LL ret=1)
{
    while(b)
    {
        if(b&1) ret=a*ret%mod;
        a=a*a%mod;
        b>>=1;
    }
    return ret;
}
LL n,m,ans;
LL inv[N],jc[N];
void init()
{
    inv[0]=jc[0]=jc[1]=1;
    for(int i=2;i<=4000000;i++) jc[i]=jc[i-1]*i%mod;
    inv[4000000]=Ksm(jc[4000000],mod-2);
    for(int i=3999999;i>=1;i--) inv[i]=inv[i+1]*(i+1)%mod;
}
LL C(LL x,LL y){return jc[y]*inv[y-x]%mod*inv[x]%mod;}
int main()
{
    read(n),read(m);
    init();
    for(int i=0;i<=m/2;i++)
    {
        LL ji=i*2+m%2,ou=m+(m-ji)/2;
        if(ji>n) break;
        ans+=C(ji,n)*C(n-1,ou+n-1)%mod;
        ans%=mod;
    }
    for(int i=m*2+1;i<=3*m;i++)
    {
        ans-=n*C(n-2,3*m-i+n-2)%mod;
        ans=(ans+mod)%mod;
    }
    printf("%lld\n",ans);
    return 0;
}
```