The value of $k$ can be very large indeed (up to $10^{12}$). Is there an efficient way to calculate the output?


Edit : 'm' is a prime number.



Calculating (n ^ fibonacci(k)) MOD m for a large value of k
Assuming $n$ and $m$ are coprime and $m$ is factored, first compute $a=\text{fibonacci}(k) \mod \varphi(m)$. Computing linear recurrence efficiently modulo $n$ is possible, e.g. via matrix exponentiation.


Then your expression is equal to $n^a \mod m$ which is easy to compute efficiently
working $\mod m$ and fast exponentiation.


$\varphi$ is [Euler totient function](https://en.wikipedia.org/wiki/Euler_totient).




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If the factorization of $m$ is unknown and $k$ is sufficiently
large, I suspect the problem is quite hard.


