A Programming Puzzle of Mode Golf Your task is to print the text Good morning, Green orb!, with every character repeated in place as many times as the most frequent byte in your source (the mode).  A trailing newline is permitted and need not be repeated.

For example if your source was

print p


Since p appears twice and every other byte appears once you would need to print

GGoooodd  mmoorrnniinngg,,  GGrreeeenn  oorrbb!!


Answers will be scored by the product of their byte count and the number of times the output is repeated.  For example the above code (if it worked) would score 7*2 = 14.  The goal should be to minimize one's score.

Your code must contain at least 1 byte.

Use this program to verify that your code and output match
Brain-Flak, 384 * 106 366 * 100 = 36,600

(((((((()()()))))))(()({}{}{}(([(({}{})){}()]((((({}())){})){}{}()(({})&lt;([{}](((({}()())))([{}]([{}]()((()()())(()()({}()([(({})){}()](((((({}))))({}({}{}{}){})(({}){}()))))&lt;((((([]){}){}){}&lt;&gt;)&lt;&gt;)&gt;[])))))))(((((()()())()){}{}){}){})&lt;&gt;(((({})&lt;&gt;)))&gt;{}{})))))&lt;(&lt;&gt;({})&lt;&gt;)&gt;))(&lt;&gt;((({}))()){}{}&lt;&gt;)&lt;&gt;(((({}&lt;&gt;()))))({}{}{}&lt;&gt;)&lt;&gt;{((&lt;&gt;{}&lt;&gt;&lt;({}&lt;&gt;)&gt;)&lt;{({}[()]&lt;(({}))&gt;)}{}&gt;)&lt;&gt;}&lt;&gt;{}


Try it online!

Explanation

The first thing I do is push the string

!bro neerG ,gninrom dooG


to the stack using pretty standard brain-flak Kolmogorov-complexity tactics.

(((((((()()()))))))(()({}{}{}(([(({}{})){}()]((((({}())){})){}{}()(({})&lt;([{}](((({}()())))([{}]([{}]()((()()())(()()({}()([(({})){}()](((((({}))))({}({}{}{}){})(({}){}()))))&lt;((((([]){}){}){}&lt;&gt;)&lt;&gt;)&gt;[])))))))(((((()()())()){}{}){}){})&lt;&gt;(((({})&lt;&gt;)))&gt;{}{})))))&lt;(&lt;&gt;({})&lt;&gt;)&gt;))(&lt;&gt;((({}))()){}{}&lt;&gt;)&lt;&gt;({}&lt;&gt;())


Then we push a counter to the off stack to tell us how many times to duplicate each character.  However I wasn't going to be able to determine what this counter was until I was done writing the program.

Next up we simultaneously reverse the string and duplicate each character in place the correct number of times.  Specifically the counter + 1.

{((&lt;&gt;{}&lt;&gt;&lt;({}&lt;&gt;)&gt;)&lt;{({}[()]&lt;(({}))&gt;)}{}&gt;)&lt;&gt;}&lt;&gt;{}


These two parts of the program have a mode of 99 open parentheses.  However since we are most certainly going to need at least 1 parenthesis.  Here is where I noticed that the last character we pushed ! conveniently has character code 33 which means we can use it to create 99, the exact number we want using only one additional parenthesis.  This is quite the coincidence but it works.