def ext_gcd(a, b):
    """
    一次不定方程式　ax + by = d(=gcd(a, b))の特殊解を求める
    :return: x, y, gcd(a, b)をタプルで返す
    """
    # a * x + b * y == d
    ret_x, ret_y, ret_gcd = 1, 0, a
    x2, y2, d2 = 0, 1, b

    while d2 != 0:
        q = ret_gcd // d2
        ret_gcd, d2 = d2, ret_gcd - d2 * q
        ret_x, x2 = x2, ret_x - x2 * q
        ret_y, y2 = y2, ret_y - y2 * q

    if ret_gcd < 0:
        ret_gcd, ret_x, ret_y = -ret_gcd, -ret_x, -ret_y

    return ret_x, ret_y, ret_gcd

def calc_inv(a, m):
    """
    ax ≡ 1 (mod m)のx(a^(-1))の値を求める
    :return: gcd(a, m),
    """
    inv, _, g = ext_gcd(a, m)

    return g, (inv + m) % m


def Chinese_Remainder_Theorem(r: list, m: list):
    """
    # x ≡ r_i (mod m_i) (0<=i<=n)を満たす連立合同式の解x (mod m0)があれば(x, m0)を返す
    # 解が存在しない場合は(None, None)が返る
    :return:
    """
    n = len(r)
    r0, m0 = 0, 1
    for i in range(n):
        assert 1 <= m[i]
        r1 = r[i] % m[i]
        m1 = m[i]

        if m0 < m1:
            r0, r1 = r1, r0
            m0, m1 = m1, m0

        if m0 % m1 == 0:
            if r0 % m1 != r1:
                return None, None
            continue

        g, im = calc_inv(m0, m1)
        u1 = m1 // g
        if (r1 - r0) % g:
            return None, None

        x = (r1 - r0) // g % u1 * im % u1
        r0 += x * m0
        m0 *= u1
        if r0 < 0:
            r0 += m0

    return r0, m0


X = []
Y = []
for i in range(3):
    x, y = map(int, input().split())
    X.append(x)
    Y.append(y)

ans, m = Chinese_Remainder_Theorem(X, Y)
print(ans if ans is not None else -1)