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 Kotlin (1.3.71) (judge verdict: Accepted)

import java.io.BufferedReader
import java.io.InputStream
import java.io.InputStreamReader
import java.io.PrintWriter
import java.lang.StringBuilder
import java.util.*

fun PrintWriter.solve(sc: FastScanner) {
    val n = sc.nextInt()
    val a = Array(n) { sc.nextLong() }.toMutableList()
    val m = (n + 1) / 2
    val left = Array(m + 1) { 0L }
    val right = Array(m + 1) { 0L }
    for (i in 0 until m) {
        left[i + 1] = left[i] + a[2 * i]
    }
    for (i in m - 1 downTo 0) {
        right[i] = right[i + 1] + if (n % 2 == 1 && i == m - 1) a[n - 1] else a[2 * i + 1]
    }
    if (n % 2 == 0) {
        var max = Long.MIN_VALUE
        for (i in 0..m) {
            val v = left[i] + right[i]
            max = Math.max(max, v)
        }
        println(max)
    } else {
        a.add(a.last())
        val maxL = Array(m + 1) { 0L }
        val maxR = Array(m + 1) { 0L }
        for (i in 0 until m) {
            maxL[i + 1] = Math.max(maxL[i] + a[2 * i + 1], left[i + 1])
        }
        for (i in m - 1 downTo 0) {
            maxR[i] = maxR[i + 1] + a[2 * i]
        }
        var max = Long.MIN_VALUE
        for (i in 0 until m) {
            val v = maxL[i] + maxR[i + 1]
            max = Math.max(max, v)
        }
        println(max)
    }
}

fun main(args: Array<String>) {
    val writer = PrintWriter(System.out, false)
    writer.solve(FastScanner(System.`in`))
    writer.flush()
}

class FastScanner(s: InputStream) {
    private var st = StringTokenizer("")
    private val br = BufferedReader(InputStreamReader(s))

    fun next(): String {
        while (!st.hasMoreTokens()) st = StringTokenizer(br.readLine())

        return st.nextToken()
    }

    fun nextInt() = next().toInt()
    fun nextLong() = next().toLong()
    fun nextLine() = br.readLine()
    fun nextDouble() = next().toDouble()
    fun ready() = br.ready()
}