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.*;
import java.io.PrintWriter;
public class Main{
	public static void main(String[] args){
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		long[][] dp=new long[n][3];
		for(int i=0; i<n; i++){
			int a=sc.nextInt();
			if(n%2==0){
				if(i<=1){
					dp[i][i]=a;
				}else if(i==2){
					dp[i][0]=dp[i-2][0]+a;
				}else{
					dp[i][0]=dp[i-2][0]+a;
					dp[i][1]=Math.max(dp[i-2][1],dp[i-3][0])+a;
				}
				System.out.println(dp[n+1][1]);
			}else{
				if(i<=1){
					dp[i][i]=a;
				}else if(i==2){
					dp[i][0]=dp[i-2][0]+a;
					dp[i][2]=a;
				}else if(i==3){
					dp[i][0]=dp[i-2][1]+a;
					dp[i][1]=dp[i-3][0]+a;
				}else{
					dp[i][0]=dp[i-2][0]+a;
					dp[i][1]=Math.max(dp[i-2][1],dp[i-3][0])+a;
					dp[i][2]=Math.max(dp[i-4][0],Math.max(dp[i-3][1],dp[i-2][2]))+a;
				}
			}
		}
		long ans=0;
		if(n%2==0){
			for(int i=n-1; i>=Math.max(0,n-2); i--){
				ans=Math.max(ans,dp[i][1-(n-1-i)]);
			}
		}else{
			for(int i=n-1; i>=Math.max(0,n-3); i--){
				ans=Math.max(ans,dp[i][2-(n-1-i)]);
			}
		}
		System.out.println(ans);
	}
}