#include <iostream>
#include <string>
#include <vector>
#include <set>
#include <cmath>
#include <algorithm>
#include <iomanip>
#include <queue>
using ll = long long;
using namespace std;

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

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

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;
}

pair<ll, ll> ChineseRem(const vector<ll> &b, const vector<ll> &m)
{
    ll r = 0, M = 1;
    for (int i = 0; i < (int)b.size(); ++i)
    {
        ll p, q;
        ll d = extGcd(M, m[i], p, q);
        if ((b[i] - r) % d != 0)
            return make_pair(-1, -1);
        ll 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(void)
{
    vector<ll> b(3), m(3);
    bool check = false;
    for (int i = 0; i < 3; ++i)
    {
        cin >> b[i] >> m[i];
        if (b[i] != 0)
            check = true;
    }
    if (!check)
    {
        ll d = Gcd(m[0], m[1]);
        ll Lcd = m[0] * m[1] / d;

        ll _d = Gcd(Lcd, m[2]);
        ll ans = Lcd * m[2] / _d;
        cout << ans << endl;
        return 0;
    }
    pair<ll, ll> ans = ChineseRem(b, m);
    cout << ans.first << endl;

    return 0;
}