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.*;

public class Main{
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] arr = new int[n + 1];
        for (int i = 1; i <= n; i++) {
            arr[i] = sc.nextInt();
        }
        long dp[][] = new long[n + 1][2];
        dp[1][1] = arr[1];
        for (int i = 2; i <= n; i++) {
            if (i % 2 == 0) {
                if (i > 2) {
                    dp[i][0] = Math.max(dp[i - 3][1], dp[i - 2][0]) + arr[i];
                } else {
                    dp[i][0] = dp[i - 2][0] + arr[i];
                }
            } else {
                dp[i][0] = Math.max(dp[i - 3][0], dp[i - 2][0]) + arr[i];
                dp[i][1] = dp[i - 2][1] + arr[i];
            }
        }
        long total;
        if (n % 2 == 0) {
            total = Math.max(dp[n - 1][1], dp[n][0]);
        } else {
            total = Math.max(Math.max(dp[n][0], dp[n - 1][0]), dp[n - 2][1]);
        }
        System.out.println(total);
    }
}