//蟻本参照
//拡張gcdやmod_invなどもある
#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <string>
#include <math.h>
#include <iomanip>
#include <limits>
#include <list>
#include <queue>
#include <tuple>
#include <map>
using namespace std;
#define MOD (long long int)(1e9+7)
#define ll long long int
#define rep(i,n) for(int i=0; i<(int)(n); i++)
#define reps(i,n) for(int i=1; i<=(int)(n); i++)
#define REP(i,n) for(int i=n-1; i>=0; i--)
#define REPS(i,n) for(int i=n; i>0; i--)
#define INF (int)(123456789)
//#define LINF (long long int)(123456789012345678)
#define LINF (long long int)(((ll)1<<31)-1)
#define all(v) v.begin(), v.end()

ll mpow(ll a, ll b){
  if(b==0) return 1;
  else if(b%2==0){ll memo = mpow(a,b/2); return memo*memo%MOD;}
  else return mpow(a,b-1) * a % MOD;
}
ll gcd(ll a, ll b){
  if(b==0) return a;
  else return gcd(b, a%b);
}
vector<ll> kaijo_memo;
ll kaijo(ll n){
  if(kaijo_memo.size() > n) return kaijo_memo[n];
  if(kaijo_memo.size() == 0) kaijo_memo.push_back(1);
  while(kaijo_memo.size() <= n) kaijo_memo.push_back(kaijo_memo[kaijo_memo.size()-1] * kaijo_memo.size() % MOD);
  return kaijo_memo[n];
}
vector<ll> gyaku_kaijo_memo;
ll gyaku_kaijo(ll n){
  if(gyaku_kaijo_memo.size() > n) return gyaku_kaijo_memo[n];
  if(gyaku_kaijo_memo.size() == 0) gyaku_kaijo_memo.push_back(1);
  while(gyaku_kaijo_memo.size() <= n) gyaku_kaijo_memo.push_back(gyaku_kaijo_memo[gyaku_kaijo_memo.size()-1] * mpow(gyaku_kaijo_memo.size(), MOD-2) % MOD);
  return gyaku_kaijo_memo[n];
}

ll nCr(ll n, ll r){
  if(n == r) return 1;//0個の丸と-1個の棒みたいな時に時に効く？不安.
  if(n < r || r < 0) return 0;
  ll ret = 1;
  ret *= kaijo(n); ret %= MOD;
  ret *= gyaku_kaijo(r); ret %= MOD;
  ret *= gyaku_kaijo(n-r); ret %= MOD;
  return ret;
}

ll extgcd(ll a, ll b, ll &x, ll &y) {
    ll g = a;
    x = 1;
    y = 0;
    if (b != 0) g = extgcd(b, a % b, y, x), y -= (a / b) * x;
    return g;
}

//aとmが互いに素でないと逆元は存在しない
ll mod_inverse(ll a, ll m){
	ll x,y;
	extgcd(a, m, x, y);
	return (m + x % m) % m;
}

//A[i] * x = B[i] (mod M[i])という連立方程式を解く
//解はb(mod m) という形になり, (b,m)を返す
pair<ll, ll> linear_congruence(const vector<ll>& A, const vector<ll>& B,
							   const vector<ll>& M){

	//最初は条件がないので全ての整数を意味するx=0 (mod 1)を解としておく
	ll x=0, m=1;

	rep(i, A.size()){
		ll a = A[i] * m, b = B[i] - A[i] * x, d = gcd(M[i], a);
		if(b%d != 0) return make_pair(0, -1);//解がない
		ll t = b/d * mod_inverse(a/d, M[i]/d) % (M[i]/d);
		x = x + m*t;
		m *= M[i]/d;
	}
	return make_pair(x%m, m);
}

int main(void){
	vector<ll> A,B,M;
	rep(i,3){
		ll x,y;
		cin>>x>>y;
		A.push_back(1);
		B.push_back(x);
		M.push_back(y);
	}
	pair<ll, ll> ans = linear_congruence(A, B, M);
	if(ans.second == -1){
		cout<<-1<<endl;
	}else{
		cout<<ans.first<<endl;
	}
	return 0;
}