# ECMAScript, NPCG+ dependent
^
(?=(x*)\1(x*))          # \1 = floor(tail/2); \2 = \1.remainder == tail%2
(?=\2*(x?))             # \3 = 1-(tail%2)
\1\2                    # tail = \1
# Assert that tail*2+\2 == N^K where N>=2 and K>=2
(\3(x*))                # \4 = floor( (potential Kth perfect root of tail*2+\2)     /2);
                        # \4.remainder = \2;
                        # \5 = floor(((potential Kth perfect root of tail*2+\2) - 1)/2);
                        # \5.remainder = \3; tail.remainder = 0
(?:                               # tail == floor((N^J-N)/2) == floor((N^(J-1)-1)*N/2);
                                  # tail.remainder == 0
    (?=
        (?=
            \4$\2                 # only needed in the case of N==2
        |
            (?=(\4\4\2)+(x*))     # \6 = \4*2+\2; \7 = remainder
            (?=.*(?=\6)\7(x*))    # \8 = \6 - \7
            \8\6*$
        )
        ((x*)(x|(?=(x))))         # \9 = floor(((N^J-N) / (\4*2+\2)  )/2)
                                  # \9.remainder = \12;
                                  # \10 = floor(((N^J-N) / (\4*2+\2)-1)/2)
                                  # \10.remainder = \11;
                                  # tail -= \9; tail.remainder = \12
        (?=
            \12
            (
                (?=\9$)               # only needed in the case of N==2
            |
                (?=(\9\9\12))         # \14 = \9*2+\12
                (?=
                    \9\9              # do first subtraction without remainder, because
                                      # tail.remainder - \9.remainder == \12 - \12 == 0
                    \14*
                    (x*)              # \15 = remainder
                )
                (?=.*(?=\14)\15(x*))  # \16 = \14 - \15
                (?=\16\14*$)
            )
        )
        (                             # \17 = tool to make tail = \9
            (?=
                .*(?=\12$)
                \3(x*)                # \18
            |
                (x)                   # \19
            )
            (?=(\18\19))              # \20 = \18+\19
            \12
            \5\19                     # tail -= \5; tail.remainder = \20
            (
                (?=.*(?=\20$)\11$)
                \10$
            |
                (?=
                    (\10\10\11)+      # \22 = \10*2+\11
                    (x*)              # \23 = remainder
                )
                (?=.*(?=\22)\20\23(x*))  # \24 = \22 - \23 - \20
                \24\22*$
            )
        )
    )
    (?=.*(?=\12$)\3$)
    \17\5                         # tail = floor((\9*2+\3 + 1 - N)/2); tail.remainder = 0
)+
$