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

        if (N%2==0) {
            // long tmp_1 = 0L;
            // long tmp_2 = 0L;
            // for (int i=0;i<N;i++) {
            //     if (i%2==0) tmp_1+=A[i];
            //     else tmp_2+=A[i];
            // }
            // System.out.println(Math.max(tmp_1, tmp_2));

            long[][] dp = new long[N][2];
            long INF = 2_000_000_000_000_000_000L;
            for (int i=0;i<N;i++) {
                for (int j=0;j<2;j++) {
                    dp[i][j] = -INF;
                }
            }
            dp[0][0] = A[0];
            dp[0][1] = 0L;
            dp[1][1] = A[1];
            for (int i=1;i<N;i++) {
                dp[i][1] = Math.max(dp[i][1], dp[i-1][0]);
                if (i-2>=0) {
                    dp[i][0] = Math.max(dp[i][0], dp[i-2][0]+A[i]);
                    dp[i][1] = Math.max(dp[i][1], dp[i-2][1]+A[i]);
                }
            }
            long ans = dp[N-1][0];
            ans = Math.max(ans, dp[N-1][1]);
            System.out.println(ans);
        } else {
            long[][] dp = new long[N][3];
            long INF = 2_000_000_000_000_000_000L;
            for (int i=0;i<N;i++) {
                for (int j=0;j<3;j++) {
                    dp[i][j] = -INF;
                }
            }
            dp[0][0] = A[0];
            dp[0][1] = 0L;
            dp[0][2] = 0L;
            dp[1][1] = A[1];
            for (int i=1;i<N;i++) {
                dp[i][1] = Math.max(dp[i][1], dp[i-1][0]);
                dp[i][2] = Math.max(dp[i][2], dp[i-1][1]);
                if (i-2>=0) {
                    dp[i][0] = Math.max(dp[i][0], dp[i-2][0]+A[i]);
                    dp[i][1] = Math.max(dp[i][1], dp[i-2][1]+A[i]);
                    dp[i][2] = Math.max(dp[i][2], dp[i-2][2]+A[i]);
                }
            }
            long ans = dp[N-1][0];
            ans = Math.max(ans, dp[N-1][1]);
            ans = Math.max(ans, dp[N-1][2]);
            System.out.println(ans);
        }
    }
}