Write a Quine in Plain English Write a quine that consists of real English words separated by single spaces. A &quot;word&quot; is defined as a string containing only lowercase and uppercase letters (/[a-zA-Z]+/ in regex). To be &quot;real&quot; your word must be recognized by the official Scrabble dictionary.
I'm using the Scrabble dictionary since it gives a definitive answer on what is and isn't valid. There are too many gray areas with a normal dictionary. Note that &quot;A&quot; and &quot;I&quot; (not to mention &quot;quine&quot;) are not valid scrabble words.
Since writing a quine only using letters and spaces is close to impossible in most programming languages, you have the option to replace the single spaces between words with a character of your choice. You also have the option to append characters to the front of the first word and the end of the last word. These added characters may be anything (including newlines and non-ASCII) except letters (a-z, A-Z). There is a penalty for adding them though (see Scoring.)
Details

As usual, the quines may not read or access their own source code. (I'd say that HQ9+'s Q command violates this.)
Output should go to stdout or a similar alternative. There is no input.
The words do not need to be capitalized correctly. They can have caps and lowercase anywhere. The sequence of words does not need to make any sense.
No word may be used more than 3 times in your program. Differently capitalized words are still the same word (e.g. 'DOG', 'dog', and 'dOg' are all the same word).
Using languages like PHP or HTML that can just cat out their contents is considered a trivial loophole and is not allowed.
The program must contain at least one word.

Scoring
Your score is the number of &quot;real words&quot; in your program plus these penalties:

+1 for every space that was replaced with another character
nn for every n characters you added before the first word (yes, that's n to the power n)
nn for every n characters you added after the last word

For example, the program
We all LIKE PROgraMmING

would score 4 because it contains 4 words; no characters were added or replaced any spaces. It's output would of course be We all LIKE PROgraMmING.
The program
!We@all LIKE#PROgraMmING- =

would score 4 + 2 + 1 + 27 = 34; 4 for the words, 2 for the replaced spaces, 1 for the ! at the front, and 27 for the - = at the end. It's output would of course be !We@all LIKE#PROgraMmING- =.
The lowest score wins. Tiebreaker goes to the answer with the fewest penalty points. If there's still a tie the highest voted answer wins.
>&lt;>, 25 words + (22 + 11 + 11) extra = 57 49

Oh god, my brain.

'brr3deep*clap6beep+orb5flap*leap4deep+clap5beep5flap*leap9deep9clap*beep+flap0placed apple alp0leap=clip?lob8blip*flip0clip.


Since the rules state "There is no input.", for this program to work you'll need to pipe in an empty file.

Oh and yes, brr is a valid Scrabble word, as are tsktsk and pfft.

Explanation

First of all, the following words are no-ops:

deep clap beep flap leap clip blip flip


This is because of two reasons:


abcdefl push numbers (10, 11, 12, 13, 14, 15 and the length of the stack respectively). In addition, i pushes -1 if EOF is met, seeing as there is no input.
The p command pops three chars v,y,x and places v at the position x,y.


Yes, >&lt;> lets you modify the source code on the fly! We don't really make use of that though, as we only need the p for popping.

If we get rid of these no-ops, we get:

'brr3*6+orb5*4+55*99*+0placed apple alp0=?lob8*0.


In a similar way, the laced app part of placed apple and the e alp part of apple alp are also no-ops, and the rr of brr just reverses the stack twice, which we can also remove:

'b3*6+orb5*4+55*99*+0pl0=?lob8*0.


Finally, something that looks like a regular >&lt;> program! The idea is to make use of the standard >&lt;> quine, which works like so:


The initial ' starts string parsing, pushing every char met until we hit another '
We keep pushing chars until the end of the line, at which point we wrap the instruction pointer back to the start (since >&lt;> is toroidal)
We land on the initial ' again, and stop string parsing. The result is that we've pushed every single char in the program (except the initial ') onto the stack.


Then the following happens:

b3*6+o                Print the initial ' quote (ASCII 39)
r                     Reverse the stack
b5*4+                 Push a ';' (ASCII 59)
55*99*+0p             Replace the 'l' of 'lob' with ';'

(print loop)
l0=?;                 If the stack is empty, terminate. Otherwise...
o                     Print the top of the stack
b8*0.                 Jump back to the beginning of the loop