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 <bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i, n) for (int i = 0; i < (n); i++)

const int N = 2e5 + 5;
ll n, a[N], f[N], sum[N];

int main() {
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        sum[i] = (i > 1 ? sum[i - 2] : 0) + a[i];
    }
    for (int i = 2; i <= n; i++) {
        if (i & 1)
            f[i] = max(f[i - 2] + a[i], f[i - 1]);
        else
            f[i] = max(f[i - 2] + a[i], sum[i - 1]);
    }
    cout << f[n] << '\n';
}
