Base-32 explanation / Manual conversion

H  e  l  l  o
48 65 6C 6C 6F

48 - 0100 1000
65 - 0110 0101
6C - 0110 1100
6C - 0110 1100
6F - 0110 1111

So we get 0100100001100101011011000110110001101111, which is 40 bits long. Num of 0's to pad to left = 40 % 5 = 0

Base32 Table:
0  00000 0
1  00001 1
2  00010 2
3  00011 3
4  00100 4
5  00101 5
6  00110 6
7  00111 7
8  01000 8
9  01001 9
10 01010 A
11 01011 B
12 01100 C
13 01101 D
14 01110 E
15 01111 F
16 10000 G
17 10001 H
18 10010 I
19 10011 J
20 10100 K
21 10101 L
22 10110 M
23 10111 N
24 11000 O
25 11001 P
26 11010 Q
27 11011 R
28 11100 S
29 11101 T
30 11110 U
31 11111 V

01001 - 9     - 9  - 9
00001 - 1     - 1  - 1
10010 - 9*2   - 18 - I
10110 - 9*2+4 - 22 - M
11000 - 6*4   - 24 - O
11011 - 6*4+3 - 27 - R
00011 - 3     - 3  - 3
01111 - 15    - 15 - F

91IMOR3F

This is how to manually encode/decode a string.

To decode a string, convert each base 32 character back into their respective 5-bit groups, count the number of bits n,  n % 4 = number of 0's to remove from left of string. Then convert that binary string back into ASCII.