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;

int main() {
  const int64_t inf = 1e18;
  int n;
  cin >> n;
  vector<int64_t> a(n + 2);
  for (int i = 0; i < n; i++) cin >> a.at(i);
  vector<vector<int64_t>> dp(n + 3, vector<int64_t>(n % 2 + 2, -inf));
  for (int i = 0; i < n % 2 + 2; i++) dp.at(i + 1).at(i) = a.at(i);
  for (int i = 1; i <= n + 2; i++) {
    for (int j = 0; j < n % 2 + 2; j++) {
      for (int k = 0; k <= j; k++) {
        if (i - 2 + k - j < 0) continue;
        dp.at(i).at(j) = max(dp.at(i).at(j), dp.at(i - 2 + k - j).at(k) + a.at(i - 1));
      }
    }
  }
  cout << dp.at(n + 2).at(n % 2 + 1) << endl;
}
