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: Accepted)

#include <bits/stdc++.h>

using namespace std;

int main()
{
  int N;
  long long A[200001];
  long long dp[200001][2];
  
  scanf("%d", &N);
  
  for (int i = 1; i <= N; i++) {
    scanf("%lld", A + i);
  }
  
  dp[0][0] = dp[0][1] = 0;
  dp[1][0] = 0;
  dp[1][1] = A[1];
  dp[2][0] = max(A[1], A[2]);
  
  long long odd_sum = A[1];
  for (int i = 3; i <= N; i++) {
    if (i % 2 == 1) {
      odd_sum += A[i];
      dp[i][1] = odd_sum;
      dp[i][0] = max({dp[i - 1][0], dp[i - 2][1], A[i] + dp[i - 2][0]});
    }
    else {
      dp[i][0] = max({dp[i - 1][1], dp[i - 2][0] + A[i], dp[i - 3][1] + A[i]});
    }
  }
  
  
  printf("%lld\n", dp[N][0]);
  
  return 0;
}