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 <iostream>
#include <vector>
#include <algorithm>
#include <math.h>

using namespace std;

int main() {

    size_t N;
    cin >> N;
    vector<long long> a(N + 1);
    vector<long long> dp(N + 1, 0);
    vector<long long> prev(N + 1, 0);
    for (size_t i = 1; i <= N; ++i) cin >> a[i];
    prev[1] = a[1];
    for (size_t i = 3; i <= N; i += 2) prev[i] = a[i] + prev[i - 2];

    for (size_t i = 2; i <= N; ++i) {
        if ((i & 1) == 0) {
            dp[i] = max(prev[i - 1], dp[i - 2] + a[i]);
        } else {
            dp[i] = max(dp[i - 1], dp[i - 2] + a[i]);
        }
    }
    
    cout << max(0ll, dp[N]);
    return 0;
}