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 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
long long n[200005]={0};
int main() 
{
    long long i,z,j,k;
    long long  sum=0,sum2=0,sum1=0;
    scanf("%lld",&z);
    k=z/2 ;
    if(z%2==0)
    {
    	for(i=0;i<z;i++)
    	scanf("%d",&n[i]);
    	for(i=0;i<z;i=i+2)
    	{
    		sum1=sum1+n[i];
		}
		for(i=1;i<z;i=i+2)
    	{
    		sum2=sum2+n[i];
		}
		if(sum1>sum2)
		  printf("%lld",sum1);
		else printf("%lld",sum2);
	}
    else
    {
    	for(i=0;i<z;i++)
    	scanf("%lld",&n[i]);
    	for(i=1;i<z;i=i+2)
    	{
    		sum2=sum2+n[i];
		}
		sum1=sum2;
		for(j=0;j<k+1;j++)
		{
			sum=0;
			for(i=0;i<z;i=i+2)
			{
				if(2*j==i)
				  continue;
			   sum=sum+n[i];
			}
			
			if(sum1<sum)
			  sum1=sum;
		}
		if(sum1>sum2)
		  printf("%lld",sum1);
		else printf("%lld",sum2);
	}
  
    return 0;
}