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 <bits/stdc++.h>
using namespace std;

int main() {
   long long n;
   cin >> n;
   vector<long long> a(n);
   for (long long i=0; i<n; i++) {
      cin >> a[i];
   }
   long long ans1 = 0;
   for (long long i=0; i<n/2; i++) {
      ans1 += a[2*i];
   }
   long long ans2 = 0;
   for (long long i=0; i<n/2; i++) {
      ans2 += a[2*i+1];
   }
   long long ans = max(ans1,ans2);
   long long ans3 = 0;
   if (n % 2 != 0) {
      ans3 = ans1 - a[0] + a[n-1];
      ans = max({ans1,ans2,ans3});
   }
   cout << ans1 << endl;
   cout << ans2 << endl;
   cout << ans3 << endl;

   cout << ans << endl;
   return 0;
}