Problem:
Score : 600 points

Problem Statement

Given is an integer sequence A_1, ..., A_N of length N.

We will choose exactly \left\lfloor \frac{N}{2} \right\rfloor elements from this sequence so that no two adjacent elements are chosen.

Find the maximum possible sum of the chosen elements.

Here \lfloor x \rfloor denotes the greatest integer not greater than x.

Constraints

2 \leq N \leq 2\times 10^5

|A_i|\leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 ... A_N

Output

Print the maximum possible sum of the chosen elements.

Sample Input 1

6
1 2 3 4 5 6

Sample Output 1

12

Choosing 2, 4, and 6 makes the sum 12, which is the maximum possible value.

Sample Input 2

5
-1000 -100 -10 0 10

Sample Output 2

0

Choosing -10 and 10 makes the sum 0, which is the maximum possible value.

Sample Input 3

10
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000

Sample Output 3

5000000000

Watch out for overflow.

Sample Input 4

27
18 -28 18 28 -45 90 -45 23 -53 60 28 -74 -71 35 -26 -62 49 -77 57 24 -70 -93 69 -99 59 57 -49

Sample Output 4

295

Solution in Rust (1.42.0) (judge verdict: Wrong Answer)

#![allow(unused_imports)]
#![allow(non_snake_case, unused)]

use std::cmp::*;
use std::collections::*;
use std::ops::*;

macro_rules! eprint {
	($($t:tt)*) => {{
		use ::std::io::Write;
		let _ = write!(::std::io::stderr(), $($t)*);
	}};
}
macro_rules! eprintln {
	() => { eprintln!(""); };
	($($t:tt)*) => {{
		use ::std::io::Write;
		let _ = writeln!(::std::io::stderr(), $($t)*);
	}};
}
macro_rules! dbg {
	($v:expr) => {{
		let val = $v;
		eprintln!("[{}:{}] {} = {:?}", file!(), line!(), stringify!($v), val);
		val
	}}
}

macro_rules! mat {
	($($e:expr),*) => { Vec::from(vec![$($e),*]) };
	($($e:expr,)*) => { Vec::from(vec![$($e),*]) };
	($e:expr; $d:expr) => { Vec::from(vec![$e; $d]) };
	($e:expr; $d:expr $(; $ds:expr)+) => { Vec::from(vec![mat![$e $(; $ds)*]; $d]) };
}

macro_rules! ok {
	($a:ident$([$i:expr])*.$f:ident()$(@$t:ident)*) => {
		$a$([$i])*.$f($($t),*)
	};
	($a:ident$([$i:expr])*.$f:ident($e:expr$(,$es:expr)*)$(@$t:ident)*) => { {
		let t = $e;
		ok!($a$([$i])*.$f($($es),*)$(@$t)*@t)
	} };
}

pub fn readln() -> String {
	let mut line = String::new();
	::std::io::stdin().read_line(&mut line).unwrap_or_else(|e| panic!("{}", e));
	line
}

macro_rules! read {
	($($t:tt),*; $n:expr) => {{
		let stdin = ::std::io::stdin();
		let ret = ::std::io::BufRead::lines(stdin.lock()).take($n).map(|line| {
			let line = line.unwrap();
			let mut it = line.split_whitespace();
			_read!(it; $($t),*)
		}).collect::<Vec<_>>();
		ret
	}};
	($($t:tt),*) => {{
		let line = readln();
		let mut it = line.split_whitespace();
		_read!(it; $($t),*)
	}};
}

macro_rules! _read {
	($it:ident; [char]) => {
		_read!($it; String).chars().collect::<Vec<_>>()
	};
	($it:ident; [u8]) => {
		Vec::from(_read!($it; String).into_bytes())
	};
	($it:ident; usize1) => {
		$it.next().unwrap_or_else(|| panic!("input mismatch")).parse::<usize>().unwrap_or_else(|e| panic!("{}", e)) - 1
	};
	($it:ident; [usize1]) => {
		$it.map(|s| s.parse::<usize>().unwrap_or_else(|e| panic!("{}", e)) - 1).collect::<Vec<_>>()
	};
	($it:ident; [$t:ty]) => {
		$it.map(|s| s.parse::<$t>().unwrap_or_else(|e| panic!("{}", e))).collect::<Vec<_>>()
	};
	($it:ident; $t:ty) => {
		$it.next().unwrap_or_else(|| panic!("input mismatch")).parse::<$t>().unwrap_or_else(|e| panic!("{}", e))
	};
	($it:ident; $($t:tt),+) => {
		($(_read!($it; $t)),*)
	};
}

pub fn main() {
	let _ = ::std::thread::Builder::new().name("run".to_string()).stack_size(32 * 1024 * 1024).spawn(run).unwrap().join();
}

const MOD: u32 = 1_000_000_007;
const INF: i64 = std::i64::MAX/2;

fn solve() {
	let n = read!(usize);
	let a = read!([i64]);
	let mut dp = vec![vec![-INF;10];n+1];
	let mut range = vec![0;n+1];
	for i in 0..n {
		range[i+1] = (i/2 + 1) as i64;
	}
	dp[1][0] = a[0];
	dp[1][1] = 0;
	dp[0][0] = 0;
	for i in 1..n {
		for k in 0..10 {
			// range[i+1] - k = c
			// range[i+1-j] - p = c
			// p = range[i+1-j] - range[i+1]
			let p = range[i] - range[i+1] - k;
			if p<0 || p>=10 {
				continue;
			}
			dp[i+1][k as usize] = dp[i][p as usize];
		}
		for j in 2..8 {
			if j>i+1 {
				break;
			}
			for k in 0..8 {
				// range[i+1] - k = c
				// range[i+1-j] - p = c - 1
				// p = range[i+1-j] - range[i+1] + k + 1
				let p = range[i+1-j] - range[i+1] + k + 1;
				if p<0 || p>=10 {
					continue;
				}
				if dp[i+1-j][p as usize]!=-INF {
					dp[i+1][k as usize] = max(dp[i+1][k as usize], dp[i+1-j][p as usize] + a[i]);
				}
			}
		}
		// for j in 0..4 {
		// 	print!("{} ", dp[i+1][j]);
		// }
		// println!();
	}
	println!("{}", dp[n][range[n] as usize- n/2]);
}


fn run() {
    solve();
}