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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <climits>
#include <ctime>
#include <cassert>
#include <numeric>
#include <functional>
#include <bitset>

using namespace std;
using lint = long long int;
long long int INF = 1001001001001001LL;
int inf = 1000000007;
long long int MOD = 1000000007LL;
double PI = 3.1415926535897932;

template<typename T1,typename T2>inline void chmin(T1 &a,const T2 &b){if(a>b) a=b;}
template<typename T1,typename T2>inline void chmax(T1 &a,const T2 &b){if(a<b) a=b;}

#define ALL(a) a.begin(),a.end()
#define RALL(a) a.rbegin(),a.rend()

/* do your best */

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

  if (n % 2 == 0) {

    lint o = 0;
    lint e = 0;
    for (int i = 0; i < n; i++) {
      if (i % 2 == 0) e += a[i];
      else o += a[i];
    }

    assert(false);
    // cout << max(e, o) << endl;
    return 0;
  }
  
  int m = n / 2;
  vector<vector<lint>> dp(m + 1, vector<lint> (3, -INF));
  dp[0][0] = 0;

  for (int i = 0; i < m; i++) {
    for (int j = 0; j < 3; j++) {
      if (dp[i][j] == -INF) continue;
      // cerr << "curState: " << i << " " << j << endl;
      for (int nj = j; nj < 3; nj++) {
        int takeId = i * 2 + nj;
        // cerr << "nextState: " << i + 1 << " " << nj << " takeId: " << takeId << endl;
        if (takeId < n) {
          dp[i + 1][nj] = max(dp[i + 1][nj], dp[i][j] + a[takeId]);
        }
      }

      // cerr << endl;
    }
  }

  lint ans = -INF;
  for (int j = 0; j < 3; j++)
    ans = max(ans, dp[m][j]);

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