Easier way to copy file into another folder using shutil

So the following code exports a csv file in the parsed folder; I need it to do the same thing in the downloads folder. So, what i did is just copied the same code, and changed the destination folder name. I know there is a way i can do this with shutil, but im unsure how.

main.py

if mylist is not None:     number = 0     filename = '/Users/poweruser/Applications/pythonwork/leadparser/parsed/newoutput%s.csv'      destination = "/Users/poweruser/downloads"     while os.path.exists(filename %  number):         number += 1         owlname = filename % number     with open('/Users/poweruser/Applications/pythonwork/leadparser/parsed/newoutput%s.csv' % number, 'w') as csvfile:         writer = csv.writer(             csvfile, delimiter=",")         writer.writerow(             ['Email', 'Website', 'Phone Number', 'Location'])         # var_dump(mylist)         for i in mylist:             writer.writerow(list(i))     # shutil.copy(owlname, destination) doesnt work as i want it too   if mylist is not None:     number = 0     filename = '/Users/poweruser/downloads/newoutput%s.csv'      while os.path.exists(filename %  number):         number += 1      with open('/Users/poweruser/downloads/newoutput%s.csv' % number, 'w') as csvfile:         writer = csv.writer(             csvfile, delimiter=",")         writer.writerow(             ['Email', 'Website', 'Phone Number', 'Location'])         # var_dump(mylist)         for i in mylist:             writer.writerow(list(i))