CF1372D.Omkar and Circle

Danny, the local Math Maniac, is fascinated by circles, Omkar's most recent creation. Help him solve this circle problem!

You are given $n$ nonnegative integers $a_1, a_2, \dots, a_n$ arranged in a circle, where $n$ must be odd (ie. $n-1$ is divisible by $2$). Formally, for all $i$ such that $2 \leq i \leq n$, the elements $a_{i - 1}$ and $a_i$ are considered to be adjacent, and $a_n$ and $a_1$ are also considered to be adjacent. In one operation, you pick a number on the circle, replace it with the sum of the two elements adjacent to it, and then delete the two adjacent elements from the circle. This is repeated until only one number remains in the circle, which we call the circular value.

Help Danny find the maximum possible circular value after some sequences of operations.

丹尼，当地的数学狂人，对奥姆卡尔最新创作的圆非常着迷。请帮助他解决这个关于圆的问题！

给你 $n$ 个非负整数 $a_1, a_2, \dots, a_n$，它们按顺时针（或逆时针）顺序排列在一个圆上，其中 $n$ 必须为奇数（即 $n-1$ 能被 $2$ 整除）。形式化地说，对所有满足 $2 \leq i \leq n$ 的 $i$，元素 $a_{i - 1}$ 和 $a_i$ 被视为相邻；此外，$a_n$ 和 $a_1$ 也被视为相邻。在一次操作中，你从圆上选择一个数，将其替换为它的两个相邻元素之和，然后将这两个相邻元素从圆中删除。重复该操作，直到圆中仅剩一个数，我们称其为圆值（circular value）。

请帮助丹尼找出经过若干次操作后所能得到的最大可能的圆值。

The first line contains one odd integer $n$ ($1 \leq n \lt 2 \cdot 10^5$, $n$ is odd) — the initial size of the circle.

The second line contains $n$ integers $a_{1},a_{2},\dots,a_{n}$ ($0 \leq a_{i} \leq 10^9$) — the initial numbers in the circle.

第一行包含一个奇数 $n$（$1 \leq n \lt 2 \cdot 10^5$，且 $n$ 为奇数）—— 圆圈的初始大小。

第二行包含 $n$ 个整数 $a_{1},a_{2},\dots,a_{n}$（$0 \leq a_{i} \leq 10^9$）—— 圆圈中的初始数字。

Output the maximum possible circular value after applying some sequence of operations to the given circle.

输出对给定圆应用若干次操作后所能得到的最大可能的环形值。

For the first test case, here's how a circular value of $17$ is obtained:

Pick the number at index $3$. The sum of adjacent elements equals $17$. Delete $7$ and $10$ from the circle and replace $2$ with $17$.

Note that the answer may not fit in a $32$-bit integer.

对于第一个测试用例，以下是得到环形值 $17$ 的过程：

选择索引为 $3$ 处的数字。相邻元素之和等于 $17$。从圆圈中删除 $7$ 和 $10$，并将 $2$ 替换为 $17$。

注意：答案可能超出 $32$ 位整数的表示范围。