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>
#include <climits>

using namespace std;
typedef long long ll;
const ll MOD = 1e9 + 7;

int main() {
    int n;
    cin >> n;
    vector<ll> a(n);
    for (int i = 0; i < n; ++i)
        cin >> a[i];
    vector<ll> f(n + 1), g(n + 1);
    f[1] = 0, g[1] = 0;
    for (int i = 2; i <= n; ++i) {
        if (i % 2 == 0) {
            f[i] = f[i - 2] + a[i - 2];
            g[i] = max(f[i - 2], g[i - 2]) + a[i - 1];
        } else {
            f[i] = max(f[i - 1], g[i - 1]);
            g[i] = max(f[i - 2], g[i - 2]) + a[i - 1];
        }
    }
    cout << max(f[n], g[n]);
}