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 Go (1.14.1) (judge verdict: Wrong Answer)

package main

import "fmt"

func main() {
        var n int
        fmt.Scan(&n)

        a := make([]int, n)
        for i := range a {
                fmt.Scanf("%d", &a[i])
        }

        can := make([]int, n)
        for i := range can {
                can[i] = i
        }

        sum := 0
        count := 0
        for count < n/2 {
                cp := can[0]

                p := 0
                c := a[cp]

                for j, i := range can {
                        if a[i] > c {
                                c = a[i]
                                p = j
                                cp = i
                        }
                }

                i := p
                if p > 0 && cp-can[p-1] == 1 {
                        i--
                }
                j := p + 1
                if j < len(can) && can[p+1]-cp == 1 {
                        j++
                }
                fmt.Println(can, c, p, cp, i, j)

                can = append(can[:i], can[j:]...)

                sum += c
                count++
        }

        fmt.Println(sum)
}