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
import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);

    int N = scanner.nextInt();
    int[] A = new int[N];
    for (int i = 0; i < N; i++) A[i] = scanner.nextInt();

    long[] csum = new long[N + 1];
    csum[1] = A[0];
    for (int i = 3; i <= N; i += 2) {
      csum[i] = csum[i - 2];
      csum[i] += A[i - 1];
    }
    long[] dp1 = new long[N + 1];
    long[] dp2 = new long[N + 1];
    for (int i = 2; i <= N; i++) {
      if (i % 2 == 0) {
        dp2[i] = Math.max(dp2[i - 2] + A[i - 1], csum[i - 1]);
      } else {
        dp1[i] = Math.max(dp1[i - 2] + A[i - 1], dp2[i - 1]);
      }
    }
    System.out.println(N % 2 == 1 ? dp1[N] : dp2[N]);
  }
}
