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 <iostream>
#include <string>
#include <cstring>
#include <algorithm>
#define ll long long int
using namespace std;

struct Data
{
	ll num;
	int index;
};

bool cmp(Data& a, Data& b)
{
	return a.num > b.num;
}

Data a[300005];
bool vis[300005];

int main(void)
{
	int len = 0;
	cin >> len;
	for (int i = 0; i < len; ++i) {
		cin >> a[i].num;
		a[i].index = i;
	}
	sort(a, a + len, cmp);
	memset(vis, true, sizeof(bool) * 300005);
	ll res = 0;
	int amount=0;
	for (int i = 0; amount!=len/2&&i<len; ++i) {
		if (vis[a[i].index]){
			res = res + a[i].num;
			if (a[i].index - 1 >= 0)
				vis[a[i].index - 1] = false;
			if (a[i].index + 1 < len)
				vis[a[i].index + 1] = false;
			vis[a[i].index] = false;
			++amount;
		}
	}
	cout << res;
	return 0;
}