/*****************

 f1+(k1-1)*d1=f2+(k2-1)*d2

 =>  (k1-1)*d1+(k2-1)*(-d2)=f2-f1;

 =>  d1*x+y*(-d2)=f2-f1   d1=>[0,n1-1]  d2=>[0,n2-1]

 求在给定范围内解的个数

 *********************/

 #include"iostream"

 #include"cstdio"

 #include"cstring"

 #include"cmath"

 #include"algorithm"

 using namespace std;

 typedef long long ll;

 void exgcd(ll n,ll m,ll &x,ll &y,ll &d)

 {

     if(!m){

         x=;y=;d=n;

     }

     else{

         exgcd(m,n%m,y,x,d);

         y-=(n/m)*x;

     }

 }

 int main()

 {

     ll f1,n1,d1,f2,n2,d2,i,j,t;

     cin>>t;

     while(t--)

     {

         ll x,y,tmp,d;

         //cin>>n1>>f1>>d1>>n2>>f2>>d2;

         scanf("%lld%lld%lld%lld%lld%lld",&n1,&f1,&d1,&n2,&f2,&d2);

         exgcd(d1,-d2,x,y,d);

         f2-=f1;

         if(f2%d){

             //cout<<"0"<<endl;

             puts("");

             continue;

         }

         x=x*(f2/d);

         y=y*(f2/d);

         //接下来  逼近[0,n1-1],[0,n2-1]

         ll t1=d2/(abs(d));

         ll t2=d1/(abs(d));

         if(x<||y<)

         {

             int i=;

             while()

             {

                 if(x+t1*i>=&&y+t2*i>=)

                     break;

                 i++;

             }

             x=x+t1*i;

             y=y+t2*i;

         }

         else

         {

             int i=;

             while()

             {

                 if(x-t1*i<||y-t2*i<)

                     break;

                 i++;

             }

             x=x-t1*(i-);

             y=y-t2*(i-);

         }

         if(x>(n1-)||y>(n2-))

         {

             //cout<<"0"<<endl;

             puts("");

             continue;

         }

         int ans=;

         while()

         {

             if(x+t1*ans>(n1-)||y+t2*ans>(n2-))

                 break;

             ans++;

         }

         cout<<ans<<endl;

         //printf("%lld\n",ans);

     }

     return ;

 }