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 <set>
#include <map>
#include <cmath>
#include <vector>
#include <string>
#include <algorithm>
#include <iomanip>
#include <queue>
#include <deque>
#include <stack>
#include <climits>
#include <math.h>
#include <cassert>
#include <unordered_map>
using namespace std;
#pragma warning(disable : 4996)
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;



int main() {
#ifdef _DEBUG
	freopen("input1.txt", "r", stdin);
	freopen("output2.txt", "w", stdout);
#endif
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	ll n;
	cin >> n;
	map<pair<ll, ll>, ll> dp;
	vector<ll> a(n + 1);
	for (ll i = 1; i <= n; i++)cin >> a[i];
	for (int i = 0; i <= 10; i++) {
		for (int j = 0; j <= 10; j++) {
			dp[{i, j}] = -1e10;
		}
	}
	dp[{1, 1}] = a[1];
	dp[{2, 1}] = max(a[1], a[2]);
	if (n & 1)dp[{3, 1}] = a[3];
	for (ll i = 3; i <= n; i++) {
		int lb = (i / 2) - 1;
		int ub = (i + 1) / 2;
		for (int j = lb; j <= ub; j++) {
			if(!(i == 3 && j == 1))dp[{i, j}] = dp[{i - 1, j}];
			dp[{i, j}] = max(dp[{i, j}], dp[{i - 2, j - 1}] + a[i]);
			dp[{i, j}] = max(dp[{i, j}], dp[{i - 3, j - 1}] + a[i]);
		}
	}
	cout << dp[{n, n / 2}];
}