CF1098E.Fedya the Potter

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 $c$ .

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 $8$ .