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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

long long n, m;
long long a[200000], b[200000], d[100001];

int main()
{
	cin >> n;

	for (int i = 0; i < n; i++) {
		cin >> a[i];

		if (i < 2)
			b[i] = a[i];
		else
			b[i] = b[i - 2] + a[i];
	}

	if (n % 2 == 0) {
		m = 0;

		for (long long i = 0; i < n; i += 2)
			m = max(m, b[i] - b[i + 1]);

		m += b[n - 1];
	}
	else {
		m = b[n - 3];

		for (int i = 0; i < n; i += 2)
			m = max(m, b[n - 1] - a[i]);

		long long k = (n - 1) / 2;
		d[k] = -1e9;

		for (long long i = k - 1; i >= 0; i--)
			d[i] = max(d[i + 1], b[i * 2 - 1] - b[i * 2]);

		for (long long i = 0; i < k - 1; i++)
			m = max(m, b[i * 2] - b[i * 2 + 1] + d[i + 1] + b[n - 1]);
	}

	cout << m << endl;
}