We define a permutation as an operation that rearranges the order of the elements $\{1,2,3,...,n\}$. There are $n!$ such permutations, one of which leaves the elements in their initial order. For $n=3$ we have $3!=6$ permutations:

• $P_1=$ keep the initial order
• $P_2=$ exchange the 1^st and 2^nd elements
• $P_3=$ exchange the 1^st and 3^rd elements
• $P_4=$ exchange the 2^nd and 3^rd elements
• $P_5=$ rotate the elements to the right
• $P_6=$ rotate the elements to the left

If we select one of these permutations, and we re-apply the same permutation repeatedly, we eventually restore the initial order.
For a permutation $P_i$, let $f(P_i)$ be the number of steps required to restore the initial order by applying the permutation $P_i$ repeatedly.
For $n=3$, we obtain:

• $f(P_1)=1$ : $(1,2,3)\to (1,2,3)$
• $f(P_2)=2$ : $(1,2,3)\to (2,1,3)\to (1,2,3)$
• $f(P_3)=2$ : $(1,2,3)\to (3,2,1)\to (1,2,3)$
• $f(P_4)=2$ : $(1,2,3)\to (1,3,2)\to (1,2,3)$
• $f(P_5)=3$ : $(1,2,3)\to (3,1,2)\to (2,3,1)\to (1,2,3)$
• $f(P_6)=3$ : $(1,2,3)\to (2,3,1)\to (3,1,2)\to (1,2,3)$

Let $g(n)$ be the average value of $f^2(P_i)$ over all permutations $P_i$ of length $n$.
$g(3)=(1^2+2^2+2^2+2^2+3^2+3^2)/3!=31/6\approx 5.166666667\mathrm{e}0$
$g(5)=2081/120\approx 1.734166667\mathrm{e}1$
$g(20)=12422728886023769167301/2432902008176640000\approx 5.106136147\mathrm{e}3$

Find $g(350)$ and write the answer in scientific notation rounded to $10$ significant digits, using a lowercase e to separate mantissa and exponent, as in the examples above.