Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

6
1 2 3 4 5 6

Sample Output 1

12

Choosing 2, 4, and 6 makes the sum 12, which is the maximum possible value.

Sample Input 2

5
-1000 -100 -10 0 10

Sample Output 2

0

Choosing -10 and 10 makes the sum 0, which is the maximum possible value.

Sample Input 3

10
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000

Sample Output 3

5000000000

Watch out for overflow.

Sample Input 4

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

Sample Output 4

295

Solution in Java (OpenJDK 1.8.0) (judge verdict: Wrong Answer)


import java.util.*;

public class Main {
    void run() {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] aArr = new int[n];

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

        long[][][] dp = new long[n + 1][n / 2 + 2][2];
        for (long[][] x: dp) {
            for (long[] x2: x) {
                Arrays.fill(x2, Integer.MIN_VALUE);
            }
        }
        dp[0][0][0] = 0;
        // dp[i][j][k] ... i番目までの整数列で、j個を使って、(直前の数を使っていなければk = 0)の最大値
//        debug(dp);
        for (int i = 0; i < n; i++) {
            for (int j = 0; j <= i / 2; j++) {
                dp[i + 1][j + 1][1] = dp[i][j][0] + aArr[i];
                dp[i + 1][j][0] = Math.max(dp[i][j][0], dp[i][j][1]);
            }
        }
//        debug(dp);
        System.out.println(Math.max(dp[n][n / 2][0], dp[n][n / 2][1]));
    }

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

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