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;
#ifdef LOCAL_DEBUG
  #include "LOCAL_DEBUG.hpp"
#endif
#define int long long

template<class T> vector<T> make_vec(size_t a) { return vector<T>(a); }
template<class T, class... Ts> auto make_vec(size_t a, Ts... ts) {
  return vector<decltype(make_vec<T>(ts...))>(a, make_vec<T>(ts...));
}
template<class T, class V>
typename enable_if<is_class<T>::value == 0>::type fill(T &t, const V &v) {
    t = v;
}
template<class T, class V>
typename enable_if<is_class<T>::value != 0>::type fill(T &t, const V &v) {
    for (auto &e : t) fill(e, v);
}
// auto v = make_vec<int>(h, w);
// fill(v, 0);
const int INF = 1LL << 60;

signed main(){

  int n; cin >> n;
  vector<int> a(n+3, 0);
  for(int i = 0; i < n; i++){
    cin >> a[i];
  }

  auto dp = make_vec<int>(n+3, 3); //ox, oxx, oxxx
  fill(dp, -INF);
  dp[0][0] = a[0];
  dp[1][1] = a[1];
  dp[2][2] = a[2];
  for(int i = 2; i < n; i++){
    dp[i][0] = dp[i-2][0] + a[i];
    if(i-3 >= 0) dp[i][1] = max(dp[i-3][0], dp[i-2][1]) + a[i];
    if(i-3 >= 0) dp[i][2] = max(dp[i-3][1], dp[i-2][2]) + a[i];
    if(i-4 >= 0) dp[i][2] = max(dp[i][2], dp[i-4][0] + a[i]);
  }
  int ans = max(dp[n-1][0], dp[n-1][1]);
  if(n%2 == 1) ans = max(ans, dp[n-1][2]);
  cout << ans << endl;

  return 0;
}