We have more than $32$ but less than $63$ characters, so we only need $6$ bits instead of $8$ bits to

represent all the characters. The first method thus reduces the size. Each character of $8$ bits is replaced by a

character of $6$ bits and all bits come directly after each other. Depending on the number of characters, this

may mean that not exactly one byte comes out at the end. We then simply fill it up with $0-$en$.$ The only case in

which we can have an ambiguity is if the last $6-$bit word extends into the next byte with $2$ bits, leaving exactly $6$

bits. Then the last $6$ bits are potentially another $6-$bit word. However, if we make sure that the $0-$bit does

not occur $6$ times as a correct character, then we can recognize the case. To do this, we simply have to

start with $000001$ instead of $000000$ to exclude $000000$ as a valid $6-$bit word. We use the bit

patterns in ascending order as above, so a is equivalent to $000001$ and so on$.$ For example, the text abc

is used as the bit pattern $011000010110001001100011$ to $000001000010000011000000,$ with

the spaces inserted after $8$ bits for readability only. In the example, $60\text{'}$s have been inserted at the end $.$ use vector. please explain the compress and decompress process in detail