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>
//#include <boost/array.hpp>
//#include <boost/range/numeric.hpp>
//#include <boost/range/algorithm.hpp>
//#include <boost/multiprecision/cpp_int.hpp>

#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

using namespace std;
//using namespace boost::multiprecision;

constexpr long long INF = LLONG_MAX;
constexpr long long MOD = 1000000007;
constexpr long double PI = M_PI;

#define FOR(i,r,n) for(ll i = (ll)(r); i < (ll)(n); ++i)
#define RFOR(i,n,r) for(ll i = (ll)(n-1); i >= r; --i)
#define REP(i,n) FOR(i,0,n)
#define RREP(i,n) RFOR(i,n,0)
#define ALL(x) x.begin(),x.end()
#define CALL(x) x.cbegin(),x.cend()
#define RALL(x) x.rbegin(),x.rend()
#define CRALL(x) x.crbegin(),x.crend()

#define cout cout << fixed << setprecision(12)
#define dbglf cerr << '\n'

#define ACCM(arr) accumulate(ALL(arr),0LL)
#define Yes(ex) cout << ((ex) ? "Yes" : "No") << endl
#define YES(ex) cout << ((ex) ? "YES" : "NO") << endl

using ll = long long int;
using ull = unsigned long long int;
using ld = long double;
using vi = vector<ll>;
using vvi = vector<vi>;
using vp = vector<pair<ll, ll>>;
using vs = vector<string>;
using vc = vector<char>;
using lst = list<ll>;
using pq = priority_queue<ll>;

template < typename T >
inline void veccin(vector<T>& v) {
    for (auto && e : v) { cin >> e; }
}

template < typename T >
inline void dbg(T a) { cerr << a << ' '; }

ll n, m, k;
ll ans = 0, sum = 0, cnt = 0;
//ld d;
string s/*1, s2, s3*/;
char c/*1, c2, c3*/;
bool flag = false;

/*--------------------template--------------------*/


int main() {
    cin >> n;
	vi a(n);
	veccin(a);
	m = n / 2;
	for(ll i = 0; i < n; i += 2){
    	ans += a[i];
    }
	if(n%2==0){
    	for(ll i = 1; i < n; i += 2){
    		sum += a[i];
    	}
    	cout << max(ans,sum) << endl;
    	return 0;
    }
	FOR(i,0,m){
    	for(ll i = 0; i < n; i += 2){
        	sum += a[i];
        }
    	ans = max(ans,sum);
    }
    cout << max(ans,sum) << endl;
}