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;

int main() {
  int N; cin >> N;
  int A[N]; for (int i = 0; i < N; i++) cin >> A[i];
  long C1[(N+1)/2+1], C2[N/2+1], max_sum[N/2+1];
  C1[0] = C2[0] = max_sum[0] = 0;
  for (int i = 0; i < N; i++) {
    if (i % 2 == 0)
      C1[i/2+1] = C1[i/2] + A[i];
    else
      C2[i/2+1] = C2[i/2] + A[i];
  }
  for (int i = 0; i < N/2; i++)
    max_sum[i+1] = max(max_sum[i] + A[2*i+1], C1[i+1]);
  if (N % 2 == 0) cout << max_sum[N/2] << endl;
  else {
    long max_s = -(1L<<60);
    for (int i = 0; i <= N/2; i++) {
      max_s = max(max_s, max_sum[i] + (C1[N/2+1] - C1[i+1]));
    }
    cout << max_s << endl;
  }
}
