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 <algorithm>
#include <vector>
#include <queue>
#include <string>

using ll = long long;

using namespace std;

#define amax(a, b) a = max(a, b)
#define INF 1e15

int main() {
    int n;
    cin >> n;
    int k = n / 2;
    ll a[n]; for (auto &ai:a) cin >> ai;
    ll dp[k + 1][3];
    fill(dp[0], dp[k+1], -INF);
    dp[0][0] = 0;
    for (int i=0; i<k; i++) {
        for (int x=0; x<3; x++) {
            for (int y=x; y<3; y++) {
                if (i*2 + y < n)
                    amax(dp[i+1][y], dp[i][x] + a[i*2 + y]);
            }
        }
    }
    cout << max({dp[k][0], dp[k][1], dp[k][2]}) << '\n';
    return 0;
}
