Problem:
Score : 900 points

Problem Statement

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.

Sample Input 1

2 2

Sample Output 1

3

After two operations, there are three possible outcomes:

x=(2,4)

x=(3,3)

x=(4,2)

For example, x=(3,3) can result after the following sequence of operations:

First, choose i=0,j=1, changing x from (0,0) to (2,1).

Second, choose i=1,j=0, changing x from (2,1) to (3,3).

Sample Input 2

3 2

Sample Output 2

19

Sample Input 3

10 10

Sample Output 3

211428932

Sample Input 4

100000 50000

Sample Output 4

3463133

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

//#include <bits/stdc++.h>
#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
//#include "boost/multiprecision/cpp_int.hpp"
//typedef boost::multiprecision::cpp_int ll;
typedef long double dd;
//#define i_7 (ll)(1E9+7)
#define i_7 998244353
#define i_5 i_7-2

ll mod(ll a){
    ll c=a%i_7;
    if(c>=0)return c;
    return c+i_7;
}
typedef pair<ll,ll> l_l;
typedef pair<dd,dd> d_d;
ll inf=(ll)1E16;
#define rep(i,l,r) for(ll i=l;i<=r;i++)
#define pb push_back
ll max(ll a,ll b){if(a<b)return b;else return a;}
ll min(ll a,ll b){if(a>b)return b;else return a;}
void Max(ll &pos,ll val){pos=max(pos,val);}//Max(dp[n],dp[n-1]);
void Min(ll &pos,ll val){pos=min(pos,val);}
void Add(ll &pos,ll val){pos=mod(pos+val);}
dd EPS=1E-9;
#define fastio ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define fi first
#define se second
#define endl "\n"
#define SORT(x) sort(x.begin(),x.end())
#define ERASE(x) x.erase(unique(x.begin(),x.end()),x.end())
#define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin())
#define POSU(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin())
//template<class T>void max(T a,T b){if(a<b)return b;else return a;}
//template<class T>void min(T a,T b){if(a>b)return b;else return a;}
//template<class T>bool Max(T&a, T b){if(a < b){a = b;return 1;}return 0;}
//template<class T>bool Min(T&a, T b){if(a > b){a = b;return 1;}return 0;}

//////////////////////////


ll po(ll i,ll p){
    if(p==0)return 1;
    else{
        i=mod(i);
        if(p==1)return i;
        if(p%2==0)return po(mod(i*i),p/2);
        return mod(i*po(i,p-1));
    }
}
ll bunbo(ll n){
    return po(n,i_5);
}


#define N 2000004
ll kai[N];
ll kai2[N];
void calc(){
    kai[0]=1;
    kai2[0]=1;
    rep(i,1,N-1){
        kai[i]=mod(kai[i-1]*i);
    }
    kai2[N-1]=po(kai[N-1],i_5);
    for(ll i=N-2;i>=0;i--){
        kai2[i]=mod(kai2[i+1]*(i+1));
    }
}
ll comb(ll n,ll k){
    if(n<k)return 0;
    if(n==0)return 1;
    return mod(mod(kai[n]*kai2[n-k])*kai2[k]);
}

int main(){fastio
    calc();
    ll n,m;cin>>n>>m;
    ll a1,a2=0,a3=0;
    a1=comb(3*m+n-1,n-1);
    rep(ma,2*m+1,3*m){
        ll res=3*m-ma;
        Add(a2,comb(res+n-2,n-2));
    }
    a2=mod(a2*n);
    rep(i,m+1,3*m){
        if((3*m-i)%2==1)continue;
        if(n<i)continue;
        ll res=(3*m-i)/2;
        Add(a3,comb(n,i)*comb(res+n-1,n-1));
    }
    cout<<mod(a1-a2-a3)<<endl;
    return 0;
}