#include #pragma GCC optimize("Ofast") using namespace std; using std::cout; using std::cin; using std::endl; using ll=long long; using ld=long double; ll ILL=2167167167167167167; const int INF=2100000000; const int mod=998244353; #define rep(i,a,b) for (ll i=a;i using _pq = priority_queue, greater>; template ll LB(vector &v,T a){return lower_bound(v.begin(),v.end(),a)-v.begin();} template ll UB(vector &v,T a){return upper_bound(v.begin(),v.end(),a)-v.begin();} template bool chmin(T &a,const T &b){if(a>b){a=b;return 1;}else return 0;} template bool chmax(T &a,const T &b){if(a void So(vector &v) {sort(v.begin(),v.end());} template void Sore(vector &v) {sort(v.begin(),v.end(),[](T x,T y){return x>y;});} void yneos(bool a){if(a) cout<<"Yes\n"; else cout<<"No\n";} template void vec_out(vector &p){for(int i=0;i<(int)(p.size());i++){if(i) cout<<" ";cout< T min(vector &a){assert(!a.empty());T ans=a[0];for(auto &x:a) chmin(ans,x);return ans;} template T max(vector &a){assert(!a.empty());T ans=a[0];for(auto &x:a) chmax(ans,x);return ans;} template T sum(vector &a){assert(!a.empty());T ans=a[0]-a[0];for(auto &x:a) ans+=x;return ans;} //a以下の素数を列挙、計算量:Nlog(log(N)) vector Eratosthenes(long long a){ if(a<2) return {}; vector p(a+1),ans; p[0]=1,p[1]=1; long long k=2; while(k*k<=a){ if(p[k]==0){ ans.push_back(k); for(long long i=2;i*k<=a;i++){ p[i*k]=1; } } k++; } while(k<=a){ if(p[k]==0) ans.push_back(k); k++; } return ans; } auto E=Eratosthenes(100'000); // return val=p(N) // a=p[0].first^p[0].second * ... *p[N-1].first^p[N-1].second // for all i: p[i].first is prime number // O(sqrt(val)) std::vector> Prime_factorization(long long val){ assert(val>=1); if(val==1){ return {}; } int ind=0; std::vector> ans; for(auto x:E){ if(x*x>val) break; if(val%x!=0) continue; ans.push_back({x,0}); while(val%x==0){ ans[ind].second++; val/=x; } ind++; } if(val!=1) ans.push_back({val,1}); return ans; } void solve(); // oddloop int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int t=1; cin>>t; rep(i,0,t) solve(); } void solve(){ vector p(4); rep(i,0,4) cin>>p[i]; auto A=Prime_factorization(p[1]-p[0]); auto B=Prime_factorization(p[3]-p[2]); set s; for(auto x:A) s.insert(x.first); for(auto x:B) s.insert(x.first); while(true){ ll add=ILL; if(p[1]-p[0]!=p[3]-p[2]){ if(p[1]-p[0]