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.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;

import java.util.*;

public class Main {
    long MOD = (long) 1e9 + 7;

    InputStream in;
    OutputStream out;

    public Main() {
        this(System.in, System.out);
    }

    public Main(InputStream in, OutputStream out) {
        this.in = in;
        this.out = out;
    }

    void run() {
        Scanner sc = new Scanner(this.in);
        PrintWriter out = new PrintWriter(this.out);

        // write your code
        // write your code
        int n = sc.nextInt();
        long[] as = new long[n];

        for (int i = 0; i < n; i++) {
            as[i] = sc.nextInt();
        }

        // dp[n][c][b] := i個目まで見て、2連続で使わないことがc回あり、直前の数を使和ない: b = 0, 使ったb = 1 時の最大値
        long[][][] dp = new long[n + 1][3][2];
        for (long[][] a : dp) for (long[] b : a) Arrays.fill(b, -(1L << 60));
        dp[1][1][0] = 0;
        dp[1][0][1] = as[0];
        for (int i = 1; i < n; i++) {
            dp[i + 1][0][0] = dp[i][0][1];
            dp[i + 1][0][1] = dp[i][0][0] + as[i];
            dp[i + 1][1][0] = Math.max(dp[i][0][0], dp[i][1][1]);
            dp[i + 1][1][1] = dp[i][1][0] + as[i];
            dp[i + 1][2][0] = Math.max(dp[i][1][0], dp[i][2][1]);
            dp[i + 1][2][1] = dp[i][2][0] + as[i];
//            debug(dp[i + 1]);
        }

        long ans = -(1L << 50);
        if (n % 2 == 0) {
            ans = Math.max(dp[n][0][0], dp[n][1][1]);
        } else {
            ans = Math.max(dp[n][2][1], dp[n][1][0]);
//            ans = Math.max(dp[n][1][0], ans);
        }
        out.println(ans);

        out.flush();
    }

    void debug(Object... os) {
        System.err.println(Arrays.deepToString(os));
    }

    public static void main(String[] args) {
        new Main().run();
    }
}