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>

int	main(void)
{
  	int	N;
  	long long	A[200002];
  	long long	i;
  	long long	m;
  	long long r;
  	long long sum0;
  	long long sum1;
  	long long sum2;
  
  	scanf("%d", &N);
  	for(i=0; i<N; i++)
    {
      	scanf("%lld", &A[i]);
    }
  
  	m = N / 2;
  	r = N % 2;
  
  	if (r == 0)
    {
      	sum0 = 0;
      	sum1 = 0;
      	for(i=0; i<m; i++)
        {
          	sum0 += A[i*2];
          	sum1 += A[i*2 + 1];
        }
      	if(sum0 > sum1)
          	printf("%lld", sum0);
     	else
          	printf("%lld", sum1);
    }
  	else
    {
      	sum0 = 0;
      	sum1 = 0;
      	sum2 = 0;
      	for(i=0; i<m; i++)
        {
          	sum0 += A[i*2];
          	sum1 += A[i*2 + 1];
          	sum2 += A[i*2 + 2];
        }
      	if(sum0 > sum1)
          	if(sum0 > sum2)
	          	printf("%lld", sum0);
      		else
              	printf("%lld", sum2);
     	else
          	if(sum1 > sum2)
	          	printf("%lld", sum1);
      		else
              	printf("%lld", sum2);
      
    }
  	return (0);
}