So q(A)=1 , q(B)=2 , …q(K)=11 , …q(Z)=26.
Let's take 2k letters, 2 of each of the first k letters of the English ABC.
Now create a string in which between each of the identical letters say each of ltr there are q(ltr) other letters. The title string and its reversal - qualify as such concatenations for k=3.
Not for every k such solutions exist (try e.g. for k=5...).
It is impossible to create such a string for k=26 (either trust me or explain why), but adding one extra symbol, say #,- which will be counted as a letter with q(#)=0 - it can be done (try for k=5 as well)
Solve for k=26 adding "#" to 2*"ABC string".