Codeforces Round 485 (Div. 2) |
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Finished |
It is the middle of 2018 and Maria Stepanovna, who lives outside Krasnokamensk (a town in Zabaikalsky region), wants to rent three displays to highlight an important problem.
There are $$$n$$$ displays placed along a road, and the $$$i$$$-th of them can display a text with font size $$$s_i$$$ only. Maria Stepanovna wants to rent such three displays with indices $$$i < j < k$$$ that the font size increases if you move along the road in a particular direction. Namely, the condition $$$s_i < s_j < s_k$$$ should be held.
The rent cost is for the $$$i$$$-th display is $$$c_i$$$. Please determine the smallest cost Maria Stepanovna should pay.
The first line contains a single integer $$$n$$$ ($$$3 \le n \le 3\,000$$$) — the number of displays.
The second line contains $$$n$$$ integers $$$s_1, s_2, \ldots, s_n$$$ ($$$1 \le s_i \le 10^9$$$) — the font sizes on the displays in the order they stand along the road.
The third line contains $$$n$$$ integers $$$c_1, c_2, \ldots, c_n$$$ ($$$1 \le c_i \le 10^8$$$) — the rent costs for each display.
If there are no three displays that satisfy the criteria, print -1. Otherwise print a single integer — the minimum total rent cost of three displays with indices $$$i < j < k$$$ such that $$$s_i < s_j < s_k$$$.
5
2 4 5 4 10
40 30 20 10 40
90
3
100 101 100
2 4 5
-1
10
1 2 3 4 5 6 7 8 9 10
10 13 11 14 15 12 13 13 18 13
33
In the first example you can, for example, choose displays $$$1$$$, $$$4$$$ and $$$5$$$, because $$$s_1 < s_4 < s_5$$$ ($$$2 < 4 < 10$$$), and the rent cost is $$$40 + 10 + 40 = 90$$$.
In the second example you can't select a valid triple of indices, so the answer is -1.
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