#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int, int>;

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 O(nlog(n))
// 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<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);  // p is inv of M/d (mod. m[i]/d)
    if ((b[i] - r) % d != 0) return make_pair(0, -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() {
  vector<ll> b(3), m(3);
  bool exist_non_zero = false;
  rep(i, 3) {
    cin >> b[i] >> m[i];
    if (b[i]) exist_non_zero = true;
  }

  pair<ll, ll> res = chineseRem(b, m);
  if (res.second == -1)
    cout << -1 << endl;
  else if (exist_non_zero)
    cout << res.first << endl;
  else
    cout << res.second << endl;
}