//yuki186.cpp
//Mon Mar 15 22:59:39 2021

#include <bits/stdc++.h>
#define INTINF 2147483647
#define LLINF 9223372036854775807
#define MOD 1000000007
#define rep(i,n) for (int i=0;i<(n);++i)

using namespace std;
using ll=long long;
typedef pair<int,int> P;

inline ll mod(ll a, ll m){
	return (a%m+m)%m;
}

ll extGCD(ll a, ll b, ll &p, ll &q){
	if (b == 0){
		p = 1;
		q = 0;
		return a;
	}
	ll d = extGCD(b, a%b, q, p);
	q -= a/b*p;
	return d;
}

//中国剰余定理。
//リターンは(r,m)。x ≡ r (mod m)
//解なしの時は(0,-1)が帰ってくる。
pair<ll,ll> ChineseRem(ll b1, ll m1, ll b2, ll m2){
	ll p,q;
	ll d = extGCD(m1,m2,p,q);
	if ((b2-b1)%d!=0) return make_pair(0,-1);
	ll m = m1*(m2/d); 
	ll tmp = (b2-b1)/d*p%(m2/d);
	ll r = mod(b1+m1*tmp,m);
	return make_pair(r,m);
}

int main(){
	vector<ll> bs(3),ms(3);
	rep(i,3) cin >> bs[i] >> ms[i];
	ll ansb = 0, ansm = 1;
	rep(i,3){
		pair<ll,ll> tmp = ChineseRem(ansb,ansm,bs[i],ms[i]);
		ansb = tmp.first;
		ansm = tmp.second;
		if (ansm==-1)break;
	}
	if (ansm==-1)cout << -1 << endl;
	else if (ansb==0) cout << ansm << endl;
	else cout << ansb << endl;
}