This question is for my cryptography class. Please convert the psuedo code (lines 1 - 9) in to python language. please make sure the output is the same as you see in figure 79. please help me and I'll give you a thumbs up

ax mod n: l: procedure FASTEXPONENTIATION(a, x) 2: x = num(xk . . . xo) for i k, k-l, 0d0 y x y mod n 4: 5: 6: 7: 8: end for 9: end procedure if Xi = l then end if Figure 79: Algorithm for fast exponentiation Squaring Multiplying 310 211 111 010 172 mod 21=16 162 mod 21 = 4 52 mod 21-4 52 mod 21 = 4 17 16 174mod 21 = 5 174mod 21 = 5 Figure 80: Computing 1722 mod 21