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;

int main() {
  int n;
  cin >> n;
  vector<ll> a(n);
  for (auto& e : a) cin >> e;

  int m = 1 + n % 2;
  vector<vector<ll>> dp(n + 2, vector<ll>(m + 1, -1e18));
  dp[0][0] = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= m; j++) {
      dp[i + 2][j] = max(dp[i + 2][j], dp[i][j] + a[i]);
      if (j + 1 <= m) dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j]);
    }
  }
  cout << max(*max_element(dp[n].begin(), dp[n].end()), *max_element(dp[n + 1].begin(), dp[n + 1].end())) << endl;
}
