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: Runtime Error)

import java.util.*
import kotlin.collections.HashMap
import kotlin.math.max

fun main() {
    val sc = Scanner(System.`in`)
    val n = sc.nextInt()
    val arr = LongArray(n+1)
    val mod = (1e9+7).toLong()

    for(i in 1..n) {
        arr[i] = sc.nextLong()
    }

    fun isValid(total:Int, remain:Int): Boolean {
        return remain * 2 - total <= 1
    }

    val dp = Array(n+1){HashMap<Int,Long>()}

    fun solve(cur:Int, remain:Int): Long {
        if(!isValid(cur, remain)) {
            return Long.MIN_VALUE
        }
        if(remain <= 0) {
            return 0L
        }
        if(dp[cur].contains(remain)) {
            return dp[cur][remain]!!
        }
        var ans = Long.MIN_VALUE
        if(isValid(cur - 1, remain)) {
            ans = max(ans, solve(cur - 1, remain))
        }
        ans = max(ans, solve(cur - 2, remain - 1) + arr[cur])
        dp[cur][remain] = ans
        return dp[cur][remain]!!
    }
    print(solve(n, n / 2))
}