#include <bits/stdc++.h>
using namespace std;
template<typename T1,typename T2>
ostream& operator << (ostream& out,pair<T1,T2> p) {
    out<<"("<<p.first<<","<<p.second<<")";
    return out;
}
template<typename T>
ostream& operator << (ostream& out,vector<T>v) {
    out<<"[";
    if(!v.empty()){
        cout<<v[0];
        for(int i=1;i<v.size();i++)cout<<","<<v[i];
    }
    out<<"]";
    return out;
}
typedef long long ll;
ll mul(ll a,ll b,ll mod){//O(1)取模快速乘，不会爆long long
	return (a*b-(ll)((long double)a/mod*b)*mod+mod)%mod;
}
ll exgcd(ll a, ll b, ll& x, ll& y){
	if(!b){
		x = 1, y = 0;
		return a;
	}
	ll d = exgcd(b,a%b,y,x);
	y -= a/b*x;
	return d;
}
pair<ll,ll> getv(ll n,vector<ll>&mo,vector<ll>&res){
	ll a1,m1;
    a1=res[0],m1=mo[0];
	bool ok = 1;
	for(ll i=1;i<n;i++){
		ll a2,m2,k1,k2;
        m2=mo[i],a2=res[i];
		ll d = exgcd(m1,m2,k1,k2);
		if((a2-a1)%d) ok = 0;
		else{
			k1=mul(k1,(a2-a1)/d,m2/d);//这个地方必须要用取模快速乘
			a1=a1+k1*m1;
			m1=abs(m1/d*m2);
		}
	}
	if(ok)return {(a1%m1+m1)%m1,m1};
    else return {-1,-1};
}
ll query(ll l,ll mod,ll k){
    return l/mod+((l%mod)>=k);
}   
ll query(ll l,ll r,ll mod,ll k){
    ll res=query(r,mod,k);
    if(l>0)res-=query(l-1,mod,k);
    return res;
}
struct kkk{
    ll x,y,mod,t;
};
ll solve()
{
    int n;cin>>n;
    ll m;cin>>m;
    vector<ll> X{0,m+1},Y{0,m+1};
    vector<kkk> a(n);
    for(int i=0;i<n;i++){
        cin>>a[i].x>>a[i].y>>a[i].mod>>a[i].t;
        X.push_back(a[i].x);
        Y.push_back(a[i].y);
    }
    sort(X.begin(),X.end());X.resize(unique(X.begin(),X.end())-X.begin());
    sort(Y.begin(),Y.end());Y.resize(unique(Y.begin(),Y.end())-Y.begin());
    ll ans=0;
    for(int xx=0;xx+1<X.size();xx++){
        ll lx=X[xx],rx=X[xx+1]-1;
        for(int yy=0;yy+1<Y.size();yy++){
            ll ly=Y[yy],ry=Y[yy+1]-1;
            map<pair<ll,ll>,vector<ll>> mo;
            map<pair<ll,ll>,vector<ll>> res;
            for(int x=0;x<60;x++){
                for(int y=0;y<60;y++){
                    int flag=0;
                    for(int i=0;i<n;i++){
                        // opx*X+opy*Y+k %mod =t
                        ll opx=1,opy=1,k=0;
                        if(lx>=a[i].x)opx=1,k+=-a[i].x;
                        else opx=-1,k+=a[i].x;
                        if(ly>=a[i].y)opy=1,k+=-a[i].y;
                        else opy=-1,k+=a[i].y;
                        ll Mod=a[i].mod;
                        if(((x*opx+y*opy+k)%Mod+Mod)%Mod==a[i].t){
                            flag++;
                        }
                    }       
                    if(flag==n)ans+=query(lx,rx,60,x)*query(ly,ry,60,y);

                }
            }
        }
    }
    return ans;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;cin>>T;while(T--)cout<<solve()<<"\n";
    return 0;
}