#include <bits/stdc++.h>
// #include <atcoder/all>
using namespace std;
// using namespace atcoder;
typedef long long ll;
#define rep(i, n) for(ll i = 0, i##_len = (n); i < i##_len; i++)
#define reps(i, s, n) for(ll i = (s), i##_len = (n); i < i##_len; i++)
#define rrep(i, n) for(ll i = (n) - 1; i >= 0; i--)
#define rreps(i, e, n) for(ll i = (n) - 1; i >= (e); i--)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) ((ll)(x).size())
#define len(x) ((ll)(x).length())
#define endl "\n"
template<class T> void chmax(T &a, const T b){ a = max(a, b); }
template<class T> void chmin(T &a, const T b){ a = min(a, b); }

struct ChineseRemainderTheorem {
private:
    static long long safe_mod(long long x, long long m) {
        x %= m;
        if (x < 0) x += m;
        return x;
    }
    
    static std::pair<long long, long long> inv_gcd(long long a, long long b) {
        a = safe_mod(a, b);
        if (a == 0) return {b, 0};
        long long s = b, t = a;
        long long m0 = 0, m1 = 1;
        while (t) {
            long long u = s / t;
            s -= t * u;
            m0 -= m1 * u;
            auto tmp = s;
            s = t;
            t = tmp;
            tmp = m0;
            m0 = m1;
            m1 = tmp;
        }
        if (m0 < 0) m0 += b / s;
        return {s, m0};
    }
    
public:
    static pair<long long, long long> build(const vector<long long>& r, const vector<long long>& m) {
        assert(r.size() == m.size());
        int n = int(r.size());
        long long r0 = 0, m0 = 1;
        for (int i = 0; i < n; i++) {
            assert(1 <= m[i]);
            long long r1 = safe_mod(r[i], m[i]), m1 = m[i];
            if (m0 < m1) {
                std::swap(r0, r1);
                std::swap(m0, m1);
            }
            if (m0 % m1 == 0) {
                if (r0 % m1 != r1) return {0, 0};
                continue;
            }
            long long g, im;
            std::tie(g, im) = inv_gcd(m0, m1);
            long long u1 = (m1 / g);
            if ((r1 - r0) % g) return {0, 0};
            long long x = (r1 - r0) / g % u1 * im % u1;
            r0 += x * m0;
            m0 *= u1;
            if (r0 < 0) r0 += m0;
        }
        return {r0, m0};
    }
};

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    // ifstream in("input.txt");
    // cin.rdbuf(in.rdbuf());
    vector<ll> x(3), y(3);
    rep(i, 3) cin >> x[i] >> y[i];
    pair<ll, ll> p = ChineseRemainderTheorem::build(x, y);
    if ((p.first == 0) && (p.second == 0)) {
        cout << -1 << endl;
    }
    else if (p.first == 0) {
        cout << p.second << endl;
    }
    else {
        cout << p.first << endl;
    }
    return 0;
}