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;

#define MAXN 200001

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int n; cin >> n;
    vector<int64_t> a(n), dp(n), os(n);
    for (auto &i: a) cin >> i;
    for (int i = 0; i < n; ++i) {
        os[i] = (i ? os[i-1] : 0) + (i % 2 ? 0 : a[i]);
    }
    for (int i = 0; i < n; ++i) {
        if (i) dp[i] = a[i] + (i > 1 ? dp[i-2] : 0);
        if (i % 2) {
            dp[i] = max(dp[i], os[i-1]);
        } else if (i){
            dp[i] = max(dp[i], dp[i-1]);
        }
    }
    cout << dp[n-1];
}