Solution

#include <iostream>
using namespace std;

int gcd(int a, int b, int &ca, int& cb)
{
    if(a == 0)
    {
        ca = 0;
        cb = 1;
        return b;
    }
    int ta, tb;
    int g = gcd(b%a, a, ta, tb);
    ca = tb - (b / a) * ta;
    cb = ta;
    return g;
}
bool diof_solve(int a, int b, int c, int& x, int& y, int &d)
{
    int xg, yg;
    d = gcd(a, b, xg, yg);
    if(c % d != 0)return false;
    x = xg * c / d;
    y = yg * c / d;
    return true;
}

int main()
{
    int rx, ry, a, b;
    cin >> rx >> ry >> a >> b;
    int x0, y0, g;
    if(!diof_solve(a, b, ry - rx, x0, y0, g))
    {
        cout << "-1\n";
        return 0;
    }
    int kl, kh;
    kl = x0*g/(-b);
    kh = y0*g/a;
    if(x0*g % b != 0)kl++;
    if(y0*g % a != 0)kh++;
    int k = max (kl, kh);
    cout << max(x0 + k * b / g - y0 + k * a / g, -1) << '\n';
    return 0;
}