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"
#include <random>
using namespace std; typedef unsigned long long _ulong; typedef long long int lint; typedef pair<lint, lint> plint; typedef pair<double long, double long> pld;
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((lint)(x).size())
#define FOR(i, begin, end) for(lint i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(lint i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
#define endk '\n'
template<class T>bool chmax(T& a, const T& b) { if (a < b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T& a, const T& b) { if (b < a) { a = b; return 1; } return 0; }
const lint MOD = 1e9 + 7, INF = 1e18;
lint dx[8] = { 0, -1, 1, 0, 1, -1, 1, -1 }, dy[8] = { 1, 0, 0, -1, -1, -1, 1, 1 };

lint N, arr[200000], sum = 0, odd_sum = 0;
lint dp[200100][5];
int main() {
	cin >> N;
	REP(i, N) {
		cin >> arr[i];
		sum += arr[i];
		if (i % 2 == 0) odd_sum += arr[i];
	}
	REP(i, N) {
		REP(j, 3) {
			dp[i][j] = -INF;
		}
	}
	if (N % 2 == 1) {
		dp[0][0] = 0;
		REP(i, N) {
			REP(j, 3) {
				REP(k, 3) {
					chmax(dp[i + 1][k + 1], dp[i][k]);
					chmax(dp[i + 2][k], dp[i][k] + arr[i]);
					chmax(dp[i + 3][k + 2], dp[i][k] + arr[i + 2]);
				}
			}
		}
		cout << dp[N + 1][2] << endk;
	}
	else {
		cout << max(odd_sum, sum - odd_sum) << endk;
	}
}