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The data in the first transmission in the sequence (`1100000`) has two 1 bits (an even number), so the parity bit is 0.
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The first transmission becomes `11000000` (or `C0` in hex).
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The data in the next transmission (`0000000`) has none (an even number again), so the parity bit is 0 again.
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The second transmission becomes `00000000` (or `00` in hex).
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The data in the next transmission (`0000000`) has zero 1 bits (an even number again), so the parity bit is 0 again.
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The second transmission thus becomes `00000000` (or `00` in hex).
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The data for the next two transmissions (`0111000` and `0001101`) have three 1 bits.
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Their parity bits are set to 1 so that they have an even number of 1 bits in the transmission.
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They are transmitted as `01110001` and `00011011` (or `71` and `1B` in hex).
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The last transmission (`1110`) has only four bits of data.
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Since exactly eight bits are transmitted at a time and the parity bit is the right most bit, three 0 bits and then the parity bit is added to make up eight bits.
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Since exactly eight bits are transmitted at a time and the parity bit is the right most bit, three 0 bits and then the parity bit are added to make up eight bits.
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It now looks like this (where `_` is the parity bit):
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```text
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1110 000_
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↑↑↑ (added 0 bits)
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```
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There is an odd number of 1 bits again, so the parity bit is to 1.
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There is an odd number of 1 bits again, so the parity bit is 1.
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The last transmission in the sequence becomes `11100001` (or `E1` in hex).
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The entire transmission sequence for this message then `11000000 00000000 01110001 00011011 11100001` (or `C0 00 71 1B E1` in hex).
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Your job is to help implement the message sequencer to calculate the transmission sequence and the decoder for receiving messages.
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The entire transmission sequence for this message is `11000000 00000000 01110001 00011011 11100001` (or `C0 00 71 1B E1` in hex).
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