Fibonacci Power

Given a number n. You are required to find ( Fib(n) ^ Fib(n) ) % 10^9 + 7, where Fib(n) is the nth fibonacci number.

Examples : 

Input : n = 4 
Output : 27 
4th fibonacci number is 3 
[ fib(4) ^ fib(4) ] % 10^9 + 7 = ( 3 ^ 3 ) % 10^9 + 7 = 27 

Input : n = 3 
Output : 4 
3th fibonacci number is 2 
[ fib(3) ^ fib(3) ] % 10^9 + 7 = ( 2 ^ 2 ) % 10^9 + 7 = 4 

If n is large, fib(n) will be huge and fib(n) ^ fib(n) is not only difficult to calculate but its storage is impossible.

Approach: 
( a ^ (p-1) ) % p = 1 where p is prime number (using Fermat Little Theorem). 
( a ^ a ) % m can be written as ( ( a % m ) ^ a ) % m 
It is also possible to write any number 'a' as a = k * ( m - 1 ) + r (Using Division Algorithm) 
where 'k' is quotient and 'r' is remainder. We can say that r = a % (m-1) 
So, Steps to reduce our calculation, lets suppose a = fib(n) 

 ( a ^ a ) % m 
= ( ( a % m ) ^ a ) % m 
= ( ( a % m ) ^ ( k * ( m - 1 ) + r ) ) % m 
= ( ( ( a % m ) ^ ( m-1 ) ) ^ k * ( a % m ) ^ r ) % m
= ( (1 ^ k) * ( a % m ) ^ r ) % m
= ( ( a % m ) ^ r ) % m 
= ( ( a % m ) ^ (a % (m-1) ) % m 

a % m and a % (m-1) are easy to calculate and easy to store. 
and we can calculate ( x ^ y ) % m using this GFG article. 
We can find n^th fibonacci number in log(n) time using this GFG article.

Below is the implementation of above approach :  

27

Time Complexity: log(n)

Auxiliary Space: O(1)