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 11.0.6) (judge verdict: Runtime Error)

import java.util.Scanner;

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();
		}
		long sum =0;
		long dp[][] = new long[3][N];
		dp[0][0] = 0;
		dp[1][0] = a[0];
		dp[0][1] = 0;
		dp[1][1] = a[1];
		dp[0][2] = 0;
		dp[1][2] = Math.max(a[0], Math.max(a[1], a[2]));
		dp[2][2] = a[0]+a[2];
		for (int i=3;i<N;i++) {
			if (i%2==0) {
				dp[0][i] = dp[0][i-2];
				dp[1][i] = Math.max(dp[1][i-2], dp[2][i-2]);
				dp[1][i] = Math.max(dp[1][i], dp[0][i-2]+a[i]);
				dp[2][i] = Math.max(dp[2][i-2]+a[i], dp[1][i-2]+a[i]);
			} else {
				dp[0][i] = dp[0][i-2]+a[i];
				dp[1][i] = Math.max(dp[2][i-2], dp[1][i-2]+a[i]);
				dp[2][i] = Math.max(dp[2][i-2], dp[1][i-2]+a[i]);
			}
		}
		if (N%2==1) {
			System.out.println(dp[2][N-1]);
		} else {
			System.out.println(Math.max(dp[2][N-1], dp[1][N-1]));
		}
	}
}