#include <iostream>
#include <string>
#include <cmath>
#include <algorithm>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <iomanip>
#define _USE_MATH_DEFINES
#include <math.h>
#include <functional>
#include <complex>
using namespace std;

#define rep(i, x) for (ll i = 0; i < x; ++i)
#define repn(i, x) for (ll i = 1; i <= x; ++i)

using ll = long long;
const ll INF = 1e17;
const ll MOD = 1000000007;
const ll MAX = 4000001;
const long double eps = 1E-14;

ll max(ll a, ll b)
{
    if (a > b)
    {
        return a;
    }
    return b;
}

ll min(ll a, ll b)
{
    if (a > b)
    {
        return b;
    }
    return a;
}

ll gcd(ll a, ll b)
{
    if (b == 0)
    {
        return a;
    }
    if (a < b)
    {
        return gcd(b, a);
    }
    return gcd(b, a % b);
}

ll lcm(ll a, ll b)
{
    return a / gcd(a, b) * b;
}

using vll = vector<ll>;
using vvll = vector<vector<ll>>;
using vvvll = vector<vector<vector<ll>>>;

///////////////////////////////////////////////////////////////////

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()
{
    vll x(3), y(3);
    rep(i, 3)
    {
        cin >> x.at(i) >> y.at(i);
    }

    pair<ll, ll> res = ChineseRem(x, y);

    if (res.second == -1)
    {
        cout << -1 << endl;
    }
    else if (res.first == 0)
    {
        cout << res.second << endl;
    }
    else
    {
        cout << res.first << endl;
    }
}