# 拡張ユークリッドの互除法を用いた解法

from math import gcd

# 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 % b == 0:
        gcd = b
        x = 1
        y = 1 - (a // b)

    else:
        gcd, pre_x, pre_y = extended_gcd(b, a % b)
        x = pre_y
        y = pre_x - (a // b) * pre_y

    return gcd, x, y


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

# 入力値チェック
assert gcd == 1
assert 1 <= A <= 3000
assert 1 <= B <= 3000
assert 0 <= a < A
assert 0 <= b < B

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