Hi, just a quick question about calculating the page size from the offset. if there are 10 bits for the offset in a 32 bit system, would the page size be 2^10 = 1kB bytes, where each offset contains 2 bits to determine which byte of the 4 bytes in the address to read, or would it be 2^10 * 4 = 4kB bytes, since each address is 4 bytes (32 / 8 = 4)?