// No.186 中華風 (Easy)
// https://yukicoder.me/problems/447

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int INF = 2147483647;
// const ll INF = 9223372036854775807;
// const ll MOD = 1e9 + 7;

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, 0) をリターン
pair<long long, long long> chinese_rem(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, 0);

		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() {
	ll X1, Y1, X2, Y2, X3, Y3;
	cin >> X1 >> Y1 >> X2 >> Y2 >> X3 >> Y3;

	pair<ll, ll> r = chinese_rem({X1, X2, X3}, {Y1, Y2, Y3});

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

	return 0;
}