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 long find_max(int i, int N, int[] A, long[] dp) {
        return  i == N-1 ? 0 : Long.max(dp[i+1], A[i] + find_max(i+2, N, A, dp));
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int N = scanner.nextInt();
        int[] A = new int[N];
        long[] dp = new long[N+1];

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

        dp[N-1] = A[N-1];
        int i = N-3;
        while(i >= 0) {
            dp[i] = dp[i+2] + A[i];
            i -= 2;
        }

        i = N-2;
        while(i >= 0) {
            dp[i] = Long.max(dp[i+1], A[i]+dp[i+2]);
            i-=2;
        }

        System.out.println(N % 2 == 0 ? dp[0] : find_max(0, N, A, dp));
    }
}
