False RSA

Roman, Serge and Andrew have decided to upgrade the famous RSA encryption algorithm. They think that the limitation that modulo $n$ used in RSA must be a product of two distinct primes is redundant. Instead they plan to use $n$ which is the product of $k$-th powers of two distinct primes: n = p^k q^k.

However, Nick pointed out that besides all other mathematical problems, the scheme may be easier to crack. That is — it is difficult to factorize $n$ which is a product of two distinct primes, because it has exactly one non-trivial factorization. On the other hand, in case n = p^k q^k there can be more different non-trivial factorizations. For example, 100 = 2^2 \cdot 5^2 has eight non-trivial factorizations: 100 = 2 \cdot 50, 100 = 2 \cdot 2 \cdot 25, 100 = 2 \cdot 2 \cdot 5 \cdot 5, 100 = 2 \cdot 5 \cdot 10, 100 = 4 \cdot 25, 100 = 4 \cdot 5 \cdot 5, 100 = 5 \cdot 20 and 100 = 10 \cdot 10.

Now Roman, Serge and Andrew wonder — given n = p^k q^k, how many different non-trivial factorizations of $n$ are there?