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: Runtime Error)

#include<stdio.h>
int func(int i,int a[],int n)
{
    int j,sum=0;
    for(i,j=1;j<n/2;i+2,j++)
        sum+=a[i];
    return sum;
}
int main()
{
    int i,count=-1,n,sum[200000],a[100000000],gsum;
    scanf("%d",&n);
    for(i=0;i<n;i++)
     scanf("%d",&a[i]);
    while(i!=n)
    {
        count+=1;
        i=count;
        sum[count]=func(i,a,n);
    }
    gsum=sum[0];
    for(i=0;i<=count;i++)
    {
        if(gsum<sum[i]);
        gsum=sum[i];
    }
    printf("%d",gsum);
    return 0;
}