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)

/*
ID: computerbox --> Huseyn Hajiyev
LANG: C++
TASK: target_mode_on
*/
#include <bits/stdc++.h>
#include <ext/pb_ds/detail/standard_policies.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <functional>
#define ordered_set tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update>
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize ("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#define _CRT_SECURE_NO_WARNINGS
//#include <boost/multiprecision/cpp_int.hpp>
//using boost::multiprecision::cpp_int;
#define FAST_READ ios_base::sync_with_stdio(0);
#define in freopen("input.txt", "r", stdin);
#define out freopen("output.txt", "w", stdout);
#define ll long long
#define debt(x,y)cout<<"#x = "<<(x)<<" and "<<"#y = "<<(y)<<endl;
#define deb(x)cout<<"#x = "<<(x)<<endl;
#define COUT(n, a) cout<< fixed << setprecision(a) << n<<endl
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define endl '\n'
#define flsh '\n'<<flush
#define arr(a,n) for(ll i=1;i<=n;i++) cout<<a[i]<<" "; cout << "\n";
#define vecc(a,n) for(ll i=0;i<n;i++) cout<<a[i]<<" "; cout << "\n";
#define CURTIME() cerr << clock() * 1.0 / CLOCKS_PER_SEC << endl
#define DTIME(ccc) __begin = clock(); ccc; cerr<<"Time of work = "<<(clock()-__begin)/CLOCKS_PER_SEC<<endl;
#define MAX3 2010
#define MAX4 20010
#define MAX5 200010
#define MAX6 2000010
#define MOD1 1000000007
#define MOD2 998244353

//cin.ignore (7777, '\n');

using namespace std;
using namespace __gnu_pbds;

#define debug(args...)(Debugger()) , args
class Debugger
{
  public:
  Debugger(const std::string& _separator=", ") :
  first(true),separator(_separator){}
  template<typename ObjectType>
  Debugger& operator,(const ObjectType& v)
  {
        if(!first)cerr<<separator;
        cerr<<v;
        first=false;
        return *this;
  }
  ~Debugger() {cerr<<endl;}      
  private:
    bool first;
    string separator;
};
 
template <typename T1, typename T2>
inline ostream& operator<<(std::ostream& os, const std::pair<T1, T2>& p)
{
   return os<<"("<<p.first<<", "<<p.second<<")";
}
template<typename T>
inline ostream &operator<<(ostream & os,const vector<T>& v)
{
    bool first=true;
    os<<"[";
    for(unsigned ll i=0;i<v.size();i++)
    {
        if(!first)os<<", ";
        os<<v[i];
        first=false;
    }
        return os<<"]";
}
template<typename T>
inline ostream &operator<<(ostream & os,const set<T>&v)
{
    bool first=true;
    os<<"[";
    for(typename set<T>::const_iterator ii=v.begin();ii!=v.end();++ii)
    {
        if(!first)os<<", ";
        os<<*ii;
        first=false;
    }
    return os<<"]";
}
template<typename T1, typename T2>
inline ostream &operator<<(ostream & os,const map<T1, T2>&v)
{
        bool first=true;
        os<<"[";
        for(typename map<T1,T2>::const_iterator ii=v.begin(); ii!=v.end();++ii)
        {
            if(!first)os<<", ";
            os<<*ii;
            first=false;
        }
        return os<<"]";
}

ll n;
ll massiv[MAX6];
ll dp[MAX6][3];


ll f(ll n)
{
  if(n==1)return 0;
  return max(max(dp[n-1][0],dp[n-1][1]),f(n-2)+massiv[n]);	
}


int main(){
FAST_READ;
cin>>n;
for(ll i=1;i<=n;i++)cin>>massiv[i];
for(ll i=2;i<=n;i+=2)
{
  dp[i][0]=dp[i-2][0]+massiv[i-1];
  dp[i][1]=max(dp[i-2][1],dp[i-2][0])+massiv[i];	
}

if(n%2==0)cout<<max(dp[n][1],dp[n][0])<<endl;
else
{
  cout<<f(n)<<endl;	
}






return 0;
}