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;
const int MAX_N = 200005;
int N, num, u, v, L, R;
long long ans, w, A[MAX_N], Answer[MAX_N], pref[MAX_N][2];
bool used[MAX_N];
priority_queue<pair<long long, pair<int, int>>> pq;
set<int> unused;
int main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin >> N;
	for (int i = 1; i <= N; i++)
	{
		cin >> A[i];
	}
	for (int i = 1; i <= N; i++)
	{
		pq.push({ A[i], {i, i + 1} });
	}
	for (int i = 0; i <= N + 2; i++)
	{
		unused.emplace(i);
	}
	for (int i = 1; i <= N + 2; i++)
	{
		pref[i][0] = pref[i - 1][0];
		pref[i][1] = pref[i - 1][1];
		pref[i][i & 1] += A[i];
	}
	while (num < N / 2)
	{
		w = pq.top().first;
		u = pq.top().second.first;
		v = pq.top().second.second;
		pq.pop();
		if (!used[u] && !used[v])
		{
			used[u] = true, used[v] = true;
			unused.erase(u);
			unused.erase(v);
			num++;
			ans += w;
			L = *prev(unused.lower_bound(u));
			R = *unused.lower_bound(v);
			pq.push({ (pref[R - 1][L & 1] - pref[L - 1][L & 1]) - (pref[R - 1][(L & 1) ^ 1] - pref[L - 1][(L & 1) ^ 1]), {L, R} });
		}
	}
	cout << ans << endl;
	return 0;
}