# aとbの最大公約数gcdと、a*x - b*y = 1 の特殊解x, y を返す
def extended_gcd(a, b):
    if a == 0:
        return b, 0, 1
    else:
        gcd, x, y = extended_gcd(b % a, a)
        return gcd, y - (b // a) * x, x


A, B, a, b = map(int, input().split())
gcd, x, y = extended_gcd(A, B)

ans = ((A * x) * (b - a) + a) % (A * B)
print(ans)