题目如下：

思路 or 题解:

抽象出数学公式：
$a+k\ast c\equiv b(mod1<<k)$
等价：
$x\ast c-y\ast (1<<k)=b-a$
用扩展欧几里得算法求解即可
答案就是：$x_0\%\frac{1<<k}{gcd}$

AC 代码如下:

#define int long long
#define ll long long
int a, b, c, k;
int exgcd(int a, int b, int &x, int &y)
{
    if (b == 0)
    {
        x = 1, y = 0;
        return a;
    }
    int d = exgcd(b, a % b, y, x);
    y = y - a / b * x;
    return d;
}
void solve()
{
    int t = 1ll << k;
    int x, y;
    int d = exgcd(c, t, x, y);
    if ((b - a) % d)
    {
        cout << "FOREVER\n";
        return;
    }
    else
    {
        x = x * (b - a) / d;
        t /= d;
        cout << (x % t + t) % t << '\n';
    }
}
signed main()
{
    buff;
    while (cin >> a >> b >> c >> k, a || b || c || k)
        solve();
}