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>
 
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> pii;
typedef tuple<ll, ll, ll> ti;
 
#define REP(a,b,c) for(ll a=(b);a<(c);a++)
#define PER(a,b,c) for(ll a=(b);a>=(c);a--)

ll ii(){ ll x; cin >> x; return x; }
string is(){ string x; cin >> x; return x; }
void oi(ll x){ cout << x; }
void od(ld x){ cout << fixed << setprecision(10) << x; }
void os(string x){ cout << x; }
void oie(ll x){ cout << x << endl; }
void ode(ld x){ cout << fixed << setprecision(10) << x << endl; }
void ose(string x){ cout << x << endl; }
void oe(){ cout << endl; }

ll f(map<ti,ll> &m, ti k, ll o){
    if(m.find(k)==m.end()){
        return LLONG_MIN;
    }
    return m[k]+o;
}

int main(){
    ll N = ii();
    vector<ll> A;
    REP(i,0,N) A.push_back(ii());

    map<ti,ll> dp;
    dp[make_tuple(0,0,0)]=0;

    REP(i,1,N+1){
        ll c=int(i/2);
        REP(j,c-1,c+1){
            dp[make_tuple(i,j,0)]=max(
             f(dp,make_tuple(i-1,j,0),0), 
             f(dp,make_tuple(i-1,j,1),0));
            dp[make_tuple(i,j,1)]=f(dp,make_tuple(i-1,j-1,0),A[i-1]);
        }
    }
    oie(max(
        dp[make_tuple(N,int(N/2),0)],
        dp[make_tuple(N,int(N/2),1)]
        ));

    return 0;
}