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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define rep2(i, m, n) for (int i = (m); i < (n); ++i)
#define rep(i, n) rep2(i, 0, n)
#define drep2(i, m, n) for (int i = (m)-1; i >= (n); --i)
#define drep(i, n) drep2(i, n, 0)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#ifdef LOCAL
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail> void debug_out(Head H, Tail... T) { cerr << ' ' << H; debug_out(T...); }
#define debug(...) cerr << 'L' << __LINE__ << " [" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#define dump(x) cerr << __LINE__ << " " << #x << " = " << (x) << endl
#else
#define debug(...) (void(0))
#define dump(x) (void(0))
#endif
template<typename T> using V = vector<T>;
template<typename T> using P = pair<T, T>;
using ll = long long;
using ld = long double;
using Vi = V<int>;  using VVi = V<Vi>;
using Vl = V<ll>;   using VVl = V<Vl>;
using Vd = V<ld>;   using VVd = V<Vd>;
using Vb = V<bool>; using VVb = V<Vb>;
using Pi = P<int>;
using Pl = P<ll>;
using Pd = P<ld>;
template<typename T> using priority_queue_rev = priority_queue<T, vector<T>, greater<T>>;
template<typename T> vector<T> make_vec(size_t n, T a) { return vector<T>(n, a); }
template<typename... Ts> auto make_vec(size_t n, Ts... ts) { return vector<decltype(make_vec(ts...))>(n, make_vec(ts...)); }
template<typename T> inline int sz(T &x) { return x.size(); }
template<typename T> inline bool chmin(T &a, T b) { if (a > b) { a = b; return true; } return false; }
template<typename T> inline bool chmax(T &a, T b) { if (a < b) { a = b; return true; } return false; }
template<typename T1, typename T2> istream &operator>>(istream &is, pair<T1, T2> &p) { is >> p.first >> p.second; return is; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p) { os << '(' << p.first << ", " << p.second << ')'; return os; }
template<typename T> istream &operator>>(istream &is, vector<T> &v) { for (auto &e : v) is >> e; return is; }
template<typename T> ostream &operator<<(ostream &os, const vector<T> &v) { for (auto &e : v) os << e << ' '; return os; }
template<typename T> inline int count_between(vector<T> &a, T l, T r) { return lower_bound(all(a), r) - lower_bound(all(a), l); } // [l, r)
inline ll cDiv(const ll x, const ll y) { return (x+y-1) / y; } // ceil(x/y)
inline int fLog2(const ll x) { assert(x > 0); return 63-__builtin_clzll(x); } // floor(log2(x))
inline int cLog2(const ll x) { assert(x > 0); return (x == 1) ? 0 : 64-__builtin_clzll(x-1); } // ceil(log2(x))
inline int popcount(const ll x) { return __builtin_popcountll(x); }
inline void fail() { cout << -1 << '\n'; exit(0); }
struct fast_ios { fast_ios(){ cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
// const int INF  = 1<<30;
const ll INFll = 1ll<<60;
// const ld EPS   = 1e-10;
// const ld PI    = acos(-1.0);
// const int MOD  = int(1e9)+7;
// const int MOD  = 998244353;


int main() {
  ll n; cin >> n;
  Vl a(n); cin >> a;

  int m = n/2;
  Vl dp(3, -INFll);
  dp[0] = 0;

  rep(i, m) {
    Vl tmp(3, -INFll);
    rep(j, 3) rep(x, 3) {
      int j2 = j + x;
      int k = 2*i + j + x;
      if (j2 >= 3) continue;
      if (k >= n) continue;
      chmax(tmp[j2], dp[j] + a[k]);
    }
    swap(dp, tmp);
    dump(dp);
  }

  ll ans = *max_element(all(dp));
  cout << ans << '\n';
}