don't care to alter your bearing excessively, so you will make close to n−1 heading changes. Thus, your way will be a polygonal chain from (0,0) to (n,n), comprising of all things considered n line portions where each section has positive integer length and vertical and level fragments substitute. Not all ways are equivalent
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You don't care to alter your bearing excessively, so you will make close to n−1 heading changes.
Thus, your way will be a polygonal chain from (0,0) to (n,n), comprising of all things considered n line portions where each section has positive integer length and vertical and level fragments substitute.
Not all ways are equivalent. You have n integers c1,c2,… ,cn where ci is the expense of the I-th section.
Utilizing these costs we can characterize the expense of the way as the amount of lengths of the portions of this way duplicated by their expense, I. e. on the off chance that the way comprises of k sections (k≤n), the expense of the way is equivalent to ∑i=1kci⋅lengthi (fragments are numbered from 1 to k in the request they are in the way).
Find the way of the base expense and print its expense.
Input
The principal line contains the single integer t (1≤t≤1000) — the number of experiments.
The principal line of each experiment contains the single integer n (2≤n≤105).
The second line of each experiment contains n integers c1,c2,… ,cn (1≤ci≤109) — the expenses of each portion.
It's dependable that the absolute amount of n doesn't surpass 105.
Output
For each experiment, print the base conceivable expense of the way from (0,0) to (n,n) comprising of all things considered n substituting sections.
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