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 long long ll;
typedef pair<int, int> P;
typedef pair<int, P> P1;
typedef pair<P, P> P2;
#define pu push
#define pb push_back
#define mp make_pair
#define eps 1e-7
#define INF 1000000000
#define mod 1000000007
#define fi first
#define sc second
#define rep(i, x) for (long long i = 0; i < x; i++)
#define repn(i, x) for (long long i = 1; i <= x; i++)
#define SORT(x) sort(x.begin(), x.end())
#define ERASE(x) x.erase(unique(x.begin(), x.end()), x.end())
#define POSL(x, v) (lower_bound(x.begin(), x.end(), v) - x.begin())
#define POSU(x, v) (upper_bound(x.begin(), x.end(), v) - x.begin())
vector<pair<string, P> > vec;
// vector<vector<int>> data(3, vector<int>(4));

map<ll, ll> m[200002];

vector<long long> A(200002);
long long N;

ll rec(ll n, ll k) {
    if (n <= 0 && k == 0) return 0;
    if ((n + 1) / 2 < k || n <= 0 || k < 0) return -9999999999999;
    if (m[n][k] != 0) return m[n][k];
    return m[n][k] = max(A[N - n] + rec(n - 2, k - 1), rec(n - 1, k));
}

int main() {
    scanf("%lld", &N);

    for (int i = 0; i <= N - 1; i++) {
        scanf("%lld", &A[i]);
    }
    ll res = 0;
    ll now = 0;
    if (N % 2 == 0) {
        now = 0;
        rep(i, N) {
            if (i % 2 == 0) now += A[i];
        }
        res = max(res, now);
        now = 0;
        rep(i, N) {
            if (i % 2 == 1) now += A[i];
        }
        res = max(res, now);
        cout << res << endl;
    } else {
        cout << rec(N, N / 2) << endl;
    }
    return 0;
}