# 拡張ユークリッドの互除法でgcd(a, b) と ax ≡ gcd(a, b) (mod b)となるxを求める
# x は a/gcd(a, b) の b/gcd(a,b) を法とした逆元ともいえる。
def gcd_inverse(a, b):
    a = a % b
    if a == 0:
        return b, 0
    s, xs = b, 0
    t, xt = a, 1
    while t:
        n = s // t
        s, t = t, s % t
        xs, xt = xt, xs - n * xt
    if xs < 0:
        xs += a // s
    return s, xs

def crt(b, m):
    assert len(b) == len(m)
    n = len(b)
    
    b0, m0 = 0, 1
    for i in range(n):
        assert 1 <= m[i]
        b1 = b[i] % m[i]
        m1 = m[i]
        if m0 < m1:
            m0, m1 = m1, m0
            b0, b1 = b1, b0
        if m0 % m1 == 0:
            # m0 = lcm(m0, m1), m1 = gcd(m0, m1) なので b0 ≡ b1 (mod m1) が解が存在する条件であり、b0が解になる
            if b0 % m1 != b1:
                return 0, 0
            continue
        # m0 * p + m1 * q = gcd(m0, m1)
        d, p = gcd_inverse(m0, m1)
        q = m1 // d
        # b0 ≡ b1 (mod gcd(m0, m1)) の不成立
        if (b1 - b0) % d:
            return 0, 0
        # x = b0 + (b1 - b0) // g * m0 * p
        b0 += (b1 - b0) // d % q * p % q * m0        
        # lcm(m0, m1)
        m0 *= q
    return b0, m0

x = [0] * 3
y = [0] * 3
for i in range(3):
    x[i], y[i] = map(int, input().split())

ans, _ = crt(x, y)
print(ans if ans else -1)