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 C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>

void swap(int *x, int *y){
  int temp;
  temp=*x;
  *x=*y;
  *y=temp;
}

int main(){
  
  int n;
  scanf("%d",&n);
  
  int a[n];
  for(int i=0; i<n; i++) scanf("%d",&a[i]);
  
  int flag;
  for(int i=n-1; i>0; i--){
    flag=0;
    for(int j=0; j<i; j++){
      if(a[j]>a[j+1]){
        swap(&a[j], &a[j+1]);
        flag=1;
      }
    }
    if(flag==0) break;
  }
  
  int ans=0;

  int k=n-1;
  for(int cnt=n/2; cnt>0; cnt--){
    ans+=a[k];
    k-=2;
  }
  
  printf("%d",ans);
  
  return 0;
}