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 kotlin.math.max

fun main() {
    val builder = StringBuilder()

    val n = readInputLine().toInt()
    val aArr = readInputLine().split(" ").map { it.toLong() }.toLongArray()
    val skipMax = 1 + n % 2

    // dp[インデックス][A_1を起点として開始すると仮定した場合のスキップ済み数][その数を用いたか0:用いていない 1:用いた]
    // 求める物はmax(dp[n - 1][skipMax][1], dp[n - 1][skipMax - 1][0])
    val dp = Array(n) { Array(skipMax + 1) { LongArray(2) { Long.MIN_VALUE } } }

    dp[0][0][1] = aArr[0]
    dp[0][1][0] = 0L

    for (i in 1 until n) {
        dp[i][0][0] = dp[i - 1][0][1]
        if (dp[i - 1][0][0] != Long.MIN_VALUE) {
            dp[i][0][1] = dp[i - 1][0][0] + aArr[i]
        }

        for (j in 1..skipMax) {
            dp[i][j][0] = max(dp[i - 1][j][1], dp[i - 1][j - 1][0])
            if (dp[i - 1][j][0] != Long.MIN_VALUE) {
                dp[i][j][1] = dp[i - 1][j][0] + aArr[i]
            }
        }
    }

    builder.appendln(max(dp[n - 1][skipMax][1], dp[n - 1][skipMax - 1][0]))

    print(builder.toString())
}

fun readInputLine(): String {
    return readLine()!!
}