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## E - Amusement Park Editorial by en_translator

We may rule out the possibility of riding the attraction with $$0$$ or less fun.

Therefore, the problem can be rephrased as follows.

You may choose $$K$$ elements out of the sequence $$B=(1,2,\dots A_1,1,2,\dots A_2, \dots, 1,2,\dots A_N)$$. What is their maximum sum?

If the length of $$B$$ is $$K$$, then simply its total sum is the answer.
We now assume that the length of $$B$$ exceeds $$K$$.

What we want to do is to sort $$B$$ and choose $$K$$ largest elements, but we have to devise a way.

Let us use binary search. Consider the following True/False question.

Are there $$K$$ or more values in $$B$$ that are at least $$m$$? $$\cdots(*)$$

$$(*)$$ is $$False$$ if $$n=1$$, and $$True$$ if $$m=2 \times10^9+1$$.
In addition, $$(*)$$ is monotonic against $$m$$.
Moreover, $$(*)$$ can be solved in an $$O(N)$$ time.

Therefore, we may do binary search against $$m$$, so we can find the minimum $$m$$ that satisfies $$(*)$$. (Let $$M$$ be the minimum $$m$$ we obtained.)

When choosing $$K$$ elements from sequence $$B$$, the following is obviously optimal:

• choose all the elements at least $$M$$
• choose $$M-1$$ for remaining elements (until the number of chosen elements reaches up to $$M$$)

Therefore, the problem has been solved.
The complexity is $$O(N \log \max(A_i))$$.

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