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.

Examples

Input

6
1 2 3 4 5 6


Output

12


Input

5
-1000 -100 -10 0 10


Output

0


Input

10
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000


Output

5000000000


Input

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


Output

295
#include <iostream>
#include <vector>

using namespace std;
typedef long long ll;

int main() {
    int N;
    cin >> N;

    vector<ll> v(N + 10);
    for (int i = 1; i <= N; i++) {
        cin >> v[i];
    }

    vector<ll> dp_fl(N + 10, 0);
    vector<ll> dp_ce(N + 10, 0);
    dp_fl[1] = 0;
    dp_ce[1] = v[1];
    dp_fl[2] = dp_ce[2] = max(v[1], v[2]);
    ll sum_of_odds = v[1];
    for (int i = 3; i <= N; i++) {
        if (i % 2 == 1) sum_of_odds += v[i];
        dp_fl[i] = max(dp_ce[i - 1], dp_fl[i - 2] + v[i]);
        dp_ce[i] = (i % 2 == 1) ? sum_of_odds : dp_fl[i];
    }
    cout << dp_fl[N] << endl;


    return 0;
}
