The first step to finding the desired number is to determine how long the number is. Obviously, a number with fewer digits is less than a number with more digits.

We can use dynamic programming to find the number of Fool's numbers with a certain length. Obviously, there are ~0~ Fool's numbers with length ~0~ or ~1~, and ~1~ Fool's number with length ~2~ or ~3~ each. With lengths greater than these, ~count(length) = count(length-2) + count(length-3)~, since a Fool's number can be made by taking an existing Fool's number and appending ~420~ or ~69~. It is enough to calculate up to ~count(125)~.

After using dynamic programming to find how long the desired Fool's number is, we now need to find the exact number, which happens to be the ~n = (N-\sum_{i=2}^{length-1} count(i))^{th}~ Fool's number with the length we found. This gets rid of the Fool's numbers which are shorter than the correct length.

Suppose that the correct Fool's number currently has a prefix ~P~; ~P~ can be empty. We note that ~P420~ is always less than ~P69~, if ~P420~ is possible. If there are enough Fool's numbers starting with ~P420~ (the exact count is ~count(length-len(P)-3)~ since there are ~length-len(P)-3~ characters remaining), the correct prefix must also start with ~P420~. Otherwise, the correct prefix would be ~P69~. If this occurs, we must now remove all ~P420~s from ~n~. Keep going until the exact Fool's number is found, that is, until ~n = 0~.

Time Complexity: ~\mathcal{O}(T \times |ans|)~, where ~|ans|~ is the length of the answer, which is less than ~125~.