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() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<long long> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (i > 1) {
      a[i] += a[i - 2];
    }
  }
  vector<long long> dp(n);
  for (int i = 1; i < n; i++) {
    if (i & 1) {
      dp[i] = a[i - 1];
    } else {
      dp[i] = dp[i - 1];
    }
    if (i > 1) {
      dp[i] = max(dp[i], dp[i - 2] + a[i] - a[i - 2]);
    } else {
      dp[i] = max(dp[i], a[i]);
    }
  }
  cout << dp[n - 1] << '\n';
  return 0;
}