Cracking $f(x) = Cx \oplus Dx$

A program I reverse engineered is using $f(x) = Cx \oplus Dx$ where C = 0x20ef138e415 and D = 0xd3eafc3af14600 as a hash function. Given a byte array, the hash is is obtained by repeatedly applying $f$ to the current hash xor next byte.

Java code:

    public static long f(long x) {
        return (0x20ef138e415L * x) ^ (0xd3eafc3af14600L * x);
    }

    public static long hash(byte[] bytes) {
        long hash = 0;

        for (byte b : bytes) {
            hash = f(hash ^ ((b & 0xff) + 1));
        }

        return f(hash);
    }

Is there an easy way to generate hash collisions?