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 C# (.NET Core 3.1.201) (judge verdict: Wrong Answer)

using System;
using System.Linq;

class F
{
	static void Main()
	{
		var n = int.Parse(Console.ReadLine());
		var a = Console.ReadLine().Split().Select(int.Parse).ToArray();

		if (n % 2 == 0)
		{
			var dp = new long[n + 5, 2];
			for (int i = 0; i <= n; i++)
				for (int j = 0; j < 2; j++)
					dp[i, j] = long.MinValue;
			dp[1, 1] = a[0];
			dp[2, 0] = a[1];

			for (int i = 0; i < n; i++)
				for (int j = 0; j < 2; j++)
				{
					if (dp[i, j] == long.MinValue) continue;
					if (i + 1 < n)
						dp[i + 2, j] = Math.Max(dp[i + 2, j], dp[i, j] + a[i + 1]);
					if (j > 0 && i + 2 < n)
						dp[i + 3, j - 1] = Math.Max(dp[i + 3, j - 1], dp[i, j] + a[i + 2]);
				}
			Console.WriteLine(Math.Max(dp[n - 1, 0], dp[n, 0]));
		}
		else
		{
			var dp = new long[n + 5, 3];
			for (int i = 0; i <= n; i++)
				for (int j = 0; j < 3; j++)
					dp[i, j] = long.MinValue;
			dp[1, 2] = a[0];
			dp[2, 1] = a[1];
			dp[3, 0] = a[2];

			for (int i = 0; i < n; i++)
				for (int j = 0; j < 3; j++)
				{
					if (dp[i, j] == long.MinValue) continue;
					if (i + 1 < n)
						dp[i + 2, j] = Math.Max(dp[i + 2, j], dp[i, j] + a[i + 1]);
					if (j > 0 && i + 2 < n)
						dp[i + 3, j - 1] = Math.Max(dp[i + 3, j - 1], dp[i, j] + a[i + 2]);
					if (j > 1 && i + 3 < n)
						dp[i + 4, j - 2] = Math.Max(dp[i + 4, j - 2], dp[i, j] + a[i + 3]);
				}
			Console.WriteLine(Math.Max(dp[n - 2, 0], Math.Max(dp[n - 1, 0], dp[n, 0])));
		}
	}
}