In part b$,$ you derived that $($N$)=$ AN$1(1),$ so writing N $1$ in the binary system, N $1=$ Pai$2$ j $,$ we get: $($N$)=$ A N$1(1)=$ A Paj$2$ j $(1)=($A $2$ j $)$ aj $(1)=...($A $4)$ a$2($A $2)$ a$1(1)$ For clarity if we want to $$nd $(23),$ it can be explained as follows: $(23)=$ A $22(1)=($A $16)1($A $8)0($A $4)1($A $2)1(1)3$ So$,$ we just need to start from computing A$2$ and multiplying it by itself to get A$4$ and again multiply by itself to get A$8$ and so on$.$ Instead of doing $22$ matrix multiplications, we only do $6$ multiplications to get A$22.$ i$)$ Implement the mentioned algorithm to rebuild your function in part d$.$is algorithm will allow us to speed up the execution time of the program. You will need to use your binary convert function in part h$.$ j$)$ Evaluate the algorithmic e$$ciency of your code in part h in the same way you did in part f$.$ I implemented the algorithm and ran it on my own computer and I have added my result below for your reference. k$)$ How long will running your code for N $=107$ take using this algorithm? Compare your $$nding with part g$.$ Algorithm ma$$ers$!$