Preparando MOJI
Fedya loves problems involving data structures. Especially ones about different queries on subsegments. Fedya had a nice array $$$a_1, a_2, \ldots a_n$$$ and a beautiful data structure. This data structure, given $$$l$$$ and $$$r$$$, $$$1 \le l \le r \le n$$$, could find the greatest integer $$$d$$$, such that $$$d$$$ divides each of $$$a_l$$$, $$$a_{l+1}$$$, ..., $$$a_{r}$$$.
Fedya really likes this data structure, so he applied it to every non-empty contiguous subarray of array $$$a$$$, put all answers into the array and sorted it. He called this array $$$b$$$. It's easy to see that array $$$b$$$ contains $$$n(n+1)/2$$$ elements.
After that, Fedya implemented another cool data structure, that allowed him to find sum $$$b_l + b_{l+1} + \ldots + b_r$$$ for given $$$l$$$ and $$$r$$$, $$$1 \le l \le r \le n(n+1)/2$$$. Surely, Fedya applied this data structure to every contiguous subarray of array $$$b$$$, called the result $$$c$$$ and sorted it. Help Fedya find the lower median of array $$$c$$$.
Recall that for a sorted array of length $$$k$$$ the lower median is an element at position $$$\lfloor \frac{k + 1}{2} \rfloor$$$, if elements of the array are enumerated starting from $$$1$$$. For example, the lower median of array $$$(1, 1, 2, 3, 6)$$$ is $$$2$$$, and the lower median of $$$(0, 17, 23, 96)$$$ is $$$17$$$.
First line contains a single integer $$$n$$$ — number of elements in array $$$a$$$ ($$$1 \le n \le 50\,000$$$).
Second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ — elements of the array ($$$1 \le a_i \le 100\,000$$$).
Print a single integer — the lower median of array $$$c$$$.
2 6 3
6
2 8 8
8
5 19 16 2 12 15
12
In the first sample array $$$b$$$ is equal to $$${3, 3, 6}$$$, then array $$$c$$$ is equal to $$${3, 3, 6, 6, 9, 12}$$$, so the lower median is $$$6$$$.
In the second sample $$$b$$$ is $$${8, 8, 8}$$$, $$$c$$$ is $$${8, 8, 8, 16, 16, 24}$$$, so the lower median is $$$8$$$.