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 <iostream>
using namespace std;
using ll = long long;
constexpr ll INF = 1LL << 60;

ll n, k, c, a[200010], dp[100010][2];
string s;

int main() {
  ios::sync_with_stdio(false); cin.tie(0);
  cin >> n;
  for(int i = 0; i < n; ++i) cin >> a[i];
  for(int i = 0; i <= n / 2 + 1; ++i) {
    dp[i][0] = -INF;
    dp[i][1] = -INF;
  }

  dp[0][0] = 0;
  for (int i = 1; i <= n; ++i) {
    int j = (i - 1) / 2;
    dp[j + 1][0] = max(dp[j + 1][0], dp[j + 1][1]);
    dp[j + 1][1] = dp[j][0] + a[i - 1];

    dp[j][0] = max(dp[j][0], dp[j][1]);
    dp[j][1] = dp[j - 1][0] + a[i - 1];
  }
  cout << max(dp[n / 2][0], dp[n / 2][1]);
  return 0;
}

