Sum my Fibonaccified divisors! The famous Fibonacci sequence is F(0) = 0; F(1) = 1; F(N+1) = F(N) + F(N-1) (for this challenge we are beginning with 0).

Your challenge: Given n, output the sum of all the dth Fibonacci numbers for all divisors d of the nth Fibonacci number. If you prefer more formal notation, 



Input: a positive integer n

Output: the sum

For example, consider n=4. F(4) = 3The divisors of 3 are 1 and 3, so the output should be F(1) + F(3) = 1 + 2 = 3.

For n=6, F(6) = 8, and the divisors of 8 are 1, 2, 4, 8, so the output is F(1) + F(2) + F(4) + F(8) = 1 + 1 + 3 + 21 = 26.

Test Cases:

1 =&gt; 1
2 =&gt; 1
3 =&gt; 2
4 =&gt; 3
5 =&gt; 6
6 =&gt; 26


This is code-golf, shortest answer in bytes wins. Standard loopholes apply.
Jelly, 7 bytes

ÆḞÆDÆḞS


Try it online!

Explanation:

ÆḞÆDÆḞS Main link (Arguments: z)
ÆḞ      zth Fibonacci number's
  ÆD                           divisors'
    ÆḞ                                   Fibonacci numbers'
      S                                                     sum