#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<ll,ll> P;
typedef pair<int,int> Pi;
#define rep(i,n) for(ll i=0;i<n;i++)
#define FOR(i,a,b) for(ll i=a;i<b;i++)
#define fi first
#define se second
#define endl "\n"

template<typename T> inline bool chmax(T &a, T b){if(a<b){a=b;return true;}return false;}
template<typename T> inline bool chmin(T &a, T b){if(a>b){a=b;return true;}return false;}
template<typename T> ostream& operator<<(ostream& s,const complex<T>& d) {return s<<"("<<d.real()<<", "<<d.imag()<< ")";}
template<typename T1, typename T2> ostream& operator<<(ostream& s,const pair<T1,T2>& d) {return s<<"("<<d.first<<", "<<d.second<<")";}
template<typename T> ostream& operator<<(ostream& s, const vector<T>& d){int len=d.size();rep(i,len){s<<d[i];if(i<len-1) s<<" ";}return s;}
template<typename T> ostream& operator<<(ostream& s,const vector<vector<T>>& d){int len=d.size();rep(i,len){s<<d[i]<<endl;}return s;}
template<typename T> ostream& operator<<(ostream& s,const set<T>& v){s<<"{ ";for(auto itr=v.begin();itr!=v.end();++itr) {if (itr!=v.begin()) {s<< ", ";}s<<(*itr);}s<<" }";return s;}
template<typename T> ostream& operator<<(ostream& s,const multiset<T>& v){s<<"{ ";for(auto itr=v.begin();itr!=v.end();++itr) {if (itr!=v.begin()) {s<< ", ";}s<<(*itr);}s<<" }";return s;}
template<typename T1, typename T2> ostream& operator<<(ostream& s,const map<T1,T2>& m){s<<"{"<<endl;for(auto itr=m.begin();itr!=m.end();++itr){s<<" "<<(*itr).first<<" : "<<(*itr).second<<endl;}s<<"}"<<endl;return s;}

const ll mod=1'000'000'007;
const ll inf=1'000'000'000'000'000'00;
const int INF=1'000'000'000;
const double EPS=1e-10;
const double PI=acos(-1);

//拡張ユークリッド
//ax+by=gcd(a,b)の解
ll extgcd(ll a,ll b,ll &x,ll &y){
	ll d=a;
	if(b){
		d=extgcd(b,a%b,y,x);
		y-=(a/b)*x;
	}else{
		x=1,y=0;
	}
	return d;
}

//中国剰余定理
//https://qiita.com/drken/items/ae02240cd1f8edfc86fd
//リターン値 (r,m) x¥equiv r (mod m)
P CRT (const vector<ll> &b,const vector<ll> &m){
	ll r=0,M=1;
	rep(i,b.size()){
		ll p,q;
		ll d=extgcd(M,m[i],p,q);
		if((b[i]-r)%d) return P(0,-1);
		ll tmp=(b[i]-r)/d*p%(m[i]/d);
		r+=M*tmp;
		M*=m[i]/d;
	}
	return P((r%M+M)%M,M);
}

int main(){
    cin.tie(0);ios::sync_with_stdio(false);
    vector<ll> b,m;
    rep(i,3){
        ll x,y;
        cin>>x>>y;
        b.push_back(x);
        m.push_back(y);
    }
    P ret=CRT(b,m);
    if(ret.se==-1) cout<<-1<<endl;
    else cout<<ret.fi<<endl;
}