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;
using ll = long long;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  ll N; cin >> N;
  ll INF = 1e15;
  vector<ll> A(N); for (ll i = 0; i < N; i++) cin >> A[i];
  vector<vector<ll>> dp(N, vector<ll>(N, -INF));

  dp[0][0] = 0;
  dp[1][1] = A[0];
  for (ll i = 1; i < N; i++) {
    vector<vector<ll>> dp2(N, vector<ll>(N, -INF));
    if (N & 1) {
      if (i & 1) {
        dp2[0][0] = max(dp[0][0], dp[0][1]);
        dp2[1][0] = max(dp[1][0], dp[1][1]);
        dp2[1][1] = dp[0][0] + A[i];
      } else {
        dp2[0][0] = max(dp[1][0], dp[1][1]);
        dp2[0][1] = dp[0][0] + A[i];
        dp2[1][1] = dp[1][0] + A[i];
      }
    } else {
      if (i & 1) {
        dp2[0][0] = dp[1][0];
        dp2[0][1] = dp[0][0] + A[i];
      } else {
        dp2[0][0] = max(dp[0][0], dp[0][1]);
        dp2[1][1] = dp[1][0] + A[i];
      }
    }
    dp = dp2;
  }

  cout << max(dp[0][0], dp[0][1]) << '\n';
  return 0;
}