e hash table is such that each bucket can potentially store 2 consecutive key-value pairs. e 15 least signi cant bits of the hash value are removed to give the bucket index [i bucket = hash(i, j, k, l)/2 15 ], and only those 15 bits need to be stored in the bucket for the key storage [hash(i, j, k, l) mod 2 ]. Hence, the key storage only requires two bytes per key, and they are sorted in increasing order, enabling a binary search within the bucket.