#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template<typename T> using V = vector<T>;
template<typename T> using VV = vector<vector<T>>;
#define fs first
#define sc second
#define rep(i,a,n) for(ll i=a;i<(ll)(n);++i)
#define ENDL '\n'
typedef pair<ll,ll> PL;


template<typename T>
T extgcd(T a,T b,T &x,T &y){
  if(!b){x=1;y=0;return a;}
  T d=extgcd(b,a%b,y,x); 
  y-=(a/b)*x;
  return d;
}
//ax+by=gcd(a,b)を満たすx,y 返り値はgcd

template<typename T>
pair<T,T> ChineseRem(const vector<T> &b,const vector<T> &m){
  T r=0,M=1;
  for(int i=0;i<(int)m.size();++i){
    T p,q;
    T g=extgcd(M,m[i],p,q);
    if((b[i]-r)%g)return {0,-1};
    r+=M*(((b[i]-r)/g*p)%(m[i]/g));
    M*=m[i]/g;
  }
  return {(r%M+M)%M,M};
}
//x=a[i](modm[i]) を満たすx(0<=x<Π(m[i])) を求める
//法が互いに素でない場合、解が存在しない場合がある



signed main(){
  cin.tie(0);ios::sync_with_stdio(false);
  cout<<fixed<<setprecision(20);
  V<ll> x(3),y(3);
  rep(i,0,3){
    cin>>x[i]>>y[i];
  }
  PL p=ChineseRem(x,y);
  if(p.sc==-1)cout<<-1<<ENDL;
  else{
    if(p.fs)cout<<p.fs<<ENDL;
    else cout<<p.sc<<ENDL;
  }
}
//( ・ __ ・ ) KEEP BEING ORGANIZED
//CHECK overflow,vector_size,what to output?