The hyperexponentiation or tetration of a number $a$ by a positive integer $b$, denoted by $a\mathbin{\uparrow \uparrow}b$ or $^b a$, is recursively defined by:

$a \mathbin{\uparrow \uparrow} 1 = a$,
$a \mathbin{\uparrow \uparrow} (k+1) = a^{(a \mathbin{\uparrow \uparrow} k)}$.

Thus we have e.g. $3 \mathbin{\uparrow \uparrow} 2 = 3^3 = 27$, hence $3 \mathbin{\uparrow \uparrow} 3 = 3^{27} = 7625597484987$ and $3 \mathbin{\uparrow \uparrow} 4$ is roughly $10^{3.6383346400240996 \cdot 10^{12}}$.

Find $(3^N \mathbin{\uparrow \uparrow} 3^N) \bmod 10^{16}$.