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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.util.*;
class Main
{
  public static void main(String args[])
  {
    long max=Long.MIN_VALUE;
    Scanner sc=new Scanner(System.in);
    int n=sc.nextInt();
    long s=0;
    long a[]=new long[n];
    for(int i=0;i<n;i++)
    {
      a[i]=sc.nextLong();
    }
    if(n%2==0)
    {
      	for(int j=0;j<2;j++)
        {
    		for(int i=j;i<n;i+=2)
       		 {
                s+=a[i];
           	 }
          max=Math.max(max,s);
          s=0;
        }
    }
    else
    {
      long sum1=0,sum2=0,sum3=0,sum4=0,sum5=0,sum6=0;
      for(int i=2;i<n-2;i+=2)
      {
        sum1=a[0]+a[i];
      }
      for(int i=3;i<n-1;i+=2)
      {
        sum2=a[0]+a[i];
      }
      for(int i=4;i<n;i+=2)
      {
        sum3=a[0]+a[i];
      }
      for(int i=3;i<n-1;i+=2)
      {
        sum4=a[1]+a[i];
      }
      for(int i=4;i<n;i+=2)
      {
        sum5=a[1]+a[i];
      }
      for(int i=4;i<n;i+=2)
      {
        sum6=a[2]+a[i];
      }
      max=Math.max(max,Math.max(sum1,Math.max(sum2,Math.max(sum3,Math.max(sum4,Math.max(sum5,sum6))))));
                                                                          }                                                                    
                                                                          
      System.out.println(max);
              
          
         
  }
}