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 MOD=1e9+7;

//const int MOD=998244353;


int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
	
	int n;
	cin>>n;
	vector<long long> v(n);
	for(auto &i:v) cin>>i;
	
	if(!(n%2)){
		long long os=0;
		long long es=0;
		for(int i=0; i<n; i++){
			if(i%2) os+=v[i];
			else es+=v[i];
		}
		cout<<max(os,es);
		return 0;
	}
	vector<long long> dp(n);
	long long os=0, es=0;
	
	for(int i=2; i<n; i+=2){
		os+=v[i-1];
		es+=v[i-2];
		dp[i]=max(os,max(es,v[i]+dp[i-2]));
	}
	cout<<dp[n-1];
	

    return 0;
}