Friendly numbers

Let $n_1, n_2, n_3, \dots, n_p$ be the positive divisors of $n$ .

Let $k_1, k_2, k_3, \dots, k_q$ be the positive divisors of $k$ .

We know that $n_1+n_2+\dots+n_{p-1}=k$ and $k_1+k_2+\dots+k_{q-1}=n$ .

Find the value of $xy$ , where

$x=\frac{1}{n_2}+\frac{1}{n_3}+\frac{1}{n_4}+\dots+\frac{1}{n_p}$

$y=\frac{1}{k_2}+\frac{1}{k_3}+\frac{1}{k_4}+\dots+\frac{1}{k_q}$

Note: $n_1=k_1=1$ and $n_p=n, k_q=k$ .