#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define i_7 (ll)(1E9+7)
#define i_5 (ll)(1E9+5)
ll mod(ll a){
    ll c=a%i_7;
    if(c>=0)return c;
    else return c+i_7;
}
typedef pair<int,int> i_i;
typedef pair<ll,ll> l_l;
ll inf=(ll)1E12;
#define rep(i,l,r) for(ll i=l;i<=r;i++)
#define pb push_back
ll max(ll a,ll b){if(a<b)return b;else return a;}
ll min(ll a,ll b){if(a>b)return b;else return a;}
void Max(ll * pos,ll val){*pos=max(*pos,val);}//Max(&dp[i][j],dp[i-1][j]);
void Min(ll * pos,ll val){*pos=min(*pos,val);}
void Add(ll * pos,ll val){*pos=mod(*pos+val);}
const long double EPS=1E-8;
////////////////////////////////////////
inline long long mod(long long a, long long m) {
    return (a % m + m) % m;
}
long long extGcd(long long a, long long b, long long &p, long long &q) {
    if (b == 0) { p = 1; q = 0; return a; }
    long long d = extGcd(b, a%b, q, p);
    q -= a/b * p;
    return d;
}
// 中国剰余定理
// リターン値を (r, m) とすると解は x ≡ r (mod. m)
// 解なしの場合は (0, -1) をリターン
pair<long long, long long> ChineseRem(const vector<long long> &b, const vector<long long> &m) {
    long long r = 0, M = 1;
    for (int i = 0; i < (int)b.size(); ++i) {
        long long p, q;
        long long d = extGcd(M, m[i], p, q); // p is inv of M/d (mod. m[i]/d)
        if ((b[i] - r) % d != 0) return make_pair(0, -1);
        long long tmp = (b[i] - r) / d * p % (m[i]/d);
        r += M * tmp;
        M *= m[i]/d;
    }
    return make_pair(mod(r, M), M);
}
///////////////////////////////////////

int main(){
    vector<ll>b,m;
    rep(i,0,2){
        ll a,c;cin>>a>>c;
        b.pb(a);m.pb(c);
    }
    l_l c=ChineseRem(b, m);
    if(c.first==0){
        cout<<-1<<endl;return 0;
    }else{
        rep(i,0,2){
            if(c.first%m[i]!=b[i]){
                cout<<-1<<endl;return 0;
            }
        }
    }
    cout<<c.first<<endl;
    return 0;
}