PythonProject: Contact book using Python смотреть онлайн

This is just me explaining the program for my School project.
the full code that I used is:
contacts = { }

def menu():
print("--------------------MENU----------------------")
print("press 4 if you want to terminate program.")
print("press 0 to look up an existing number.")
print("press 1 to add a Person to the contact book.")
print("press 2 to update an existing number")
print("press 3 to delete an existing number")
print("press 5 to print all contacts")
print("press any number from 6-9 to display Menu")
print("----------------------------------------------")
print(" DIGITAL CONTACT BOOK ")
menu()
def display():
for key, value1 in contacts.items():
print(key, ":", value1)

while True:
choice = int(input("Enter command:"))

if choice == 0:
look = input("Name of number owner:").upper()
keys = list(contacts.keys())
values1 = list(contacts.values())

print("Number of ", look, ":", values1[keys.index(look)])
elif choice == 1:
Name = input("Name of new contact: ").upper()
Number = input("Enter 11 or 8 digit number: ")
if len(Number) == 11:
contacts.update({Name: [Number, "Mobile"]})
print(" CONTACTS")
display()

elif len(Number) == 8:
contacts.update({Name:[Number, "Landline"]})
print(" CONTACTS")
display()
else:
print("Number is invalid.Please try again.")

elif choice == 2:
Name = input("Name of person to update contact:").upper()
Number = input("New contact number( 11 or 8 digits):")
if len(Number) == 11:
contacts[Name] = [Number, "Mobile"]
print(" CONTACTS")
display()

elif len(Number) == 8:
contacts[Name] = [Number, "Landline"]
print(" CONTACTS")
display()
else:
print("Number is invalid.Please try again.")

elif choice == 3:
delete = input("Name of contact to remove:").upper()
del contacts[delete]
print(" CONTACTS")
display()
elif choice == 4:
print("Thank you for using the Digital Contact Book!")
break
elif choice == 5:
print(" CONTACTS")
display()
else:
menu()