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: Runtime Error)

#include <bits/stdc++.h>
#define rep(i, n) for (lli i = 0; i < (n); i++)
#define rrep(i, n) for (lli i = (n)-1; i >= 0; i--)
#define vall(x) (x).begin(), (x).end()
using namespace std;
using lli = long long int;
void YESNO(bool), YesNo(bool);
template <class T1, class T2>
bool chmin(T1 &l, const T2 &r);
template <class T1, class T2>
bool chmax(T1 &l, const T2 &r);
template <class T1, class T2>
void vadd(vector<T1> &v, T2 x);

#define int long long int
lli dp[100005][2][4] = {};
void solve(long long N, std::vector<long long> A)
{
  rep(i, N + 1) rep(j, 2) rep(k, 4) dp[i][j][k] = -1e15;
  dp[0][0][0] = 0;
  dp[0][1][0] = 0;

  rep(i, N) rep(k, 3)
  {
    // skip
    chmax(dp[i + 1][1][k], dp[i][0][k]);
    chmax(dp[i + 1][1][k + 1], dp[i][1][k]);

    // not skip
    chmax(dp[i + 1][0][k], dp[i][1][k] + A[i]);
  }
  if (N % 2 == 0)
  {
    cout << max({dp[N][0][0], dp[N][0][1], dp[N][1][1], dp[N][1][1]}) << endl;
  }
  else
  {
    // 1回飛ばした
    // rep(i, 2) rep(j, 3)
    // {
    //   cout << dp[N][i][j] << " ";
    // }
    cout << max({dp[N][0][0], dp[N][0][1], dp[N][1][1], dp[N][1][1], dp[N][0][2]}) << endl;
  }
}

signed main()
{
  long long N;
  scanf("%lld", &N);
  std::vector<long long> A(N);
  for (int i = 0; i < N; i++)
  {
    scanf("%lld", &A[i]);
  }
  solve(N, std::move(A));
  return 0;
}

// -- lib
void YESNO(bool b) { cout << (b ? "YES" : "NO") << endl; }
void YesNo(bool b) { cout << (b ? "Yes" : "No") << endl; }

template <class T1, class T2>
bool chmin(T1 &l, const T2 &r)
{
  return (l > r) ? (l = r, true) : false;
}

template <class T1, class T2>
bool chmax(T1 &l, const T2 &r)
{
  return (l < r) ? (l = r, true) : false;
}

template <class T1, class T2>
void vadd(vector<T1> &v, T2 x)
{
  for (auto &s : v)
    s += T2(x);
}