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)

//OPTIMIZATONS
#pragma GCC optimize("02")
//CODE
#include<bits/stdc++.h>
// uncomment the below code to use int as long long
#define int long long
// uncomment the below code to use float as double
//#define float double
#define endl "\n"
#define debug cout<<"Good uptill here"<<endl<<flush;
#define deb(x) cout<<#x<<":"<<x<<endl<<flush;

using namespace std;
const int inf = 9e18;
const int N = 2e5 + 1;
int n;
int arr[N];
int dp[N][3];

int func(int a,int b){
	if(a>=n){
		if(b<1) return -inf;
		return 0;
	}
	if(dp[a][b]!=inf) return dp[a][b];
	if(n%2==0){ 
		if(b==1) dp[a][b] = func(a+2,b)+arr[a];
		else dp[a][b] = max(func(a+1,b+1),func(a+2,b)+arr[a]);
	}else{
		if(b==2) dp[a][b] = func(a+2,b)+arr[a];
		else dp[a][b] = max(func(a+1,b+1),func(a+2,b)+arr[a]);
	}
	return dp[a][b];
}

int32_t main(){
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);
	//CODE FOR SPECIFIED INPUT AND OUTPUT
	//freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);
	for(int i=0;i<N;i++)for(int j=0;j<3LL;j++) dp[i][j]=inf;
	cin>>n;
	for(int i=0;i<n;i++) cin>>arr[i];
	int ans = func(0,0);
	cout<<ans<<endl;
	return 0;
}