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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define REP(i,n) for(int i=0;i<(n);++i)
#define ALL(v)   (v).begin(),(v).end()
#define debug(x) cerr<<#x<<": "<<(x)<<endl
#define INF      (int)1e9
#define EPS      (double)1e-9
#define MOD      ((int)1e9+7)
using namespace std;
typedef long long     llong;
typedef vector<int>   vi;
typedef vector<vi >   vvi;
typedef vector<vvi >  vvvi;
typedef pair<int,int> pii;
template<class Type> void line(const Type &a){int cnt=0;for(const auto &elem:a){if(cnt++)cout<<' ';cout<<elem;}cout<<endl;}

//解説参考.
int main(){
	int n;
	cin>>n;
	
	vector<llong> a(n);
	REP(i,n) cin>>a[i];
	
	vector<vector<llong> > dp(n,vector<llong>(n+1,-INF));
	dp[0][0]=0;
	dp[0][1]=a[0];
	dp[1][0]=0;
	dp[1][1]=max(a[0],a[1]);
	for(int i=2;i<n;++i){
		for(int j=(i+1-1)/2;j<=(i+1+1)/2;++j){
			dp[i][j]=max(dp[i-1][j],dp[i-2][j-1]+a[i]);
		}
	}
	
	//REP(i,n) cout<<dp[i][(i+1)/2]<<endl;
	
	cout<<dp[n-1][n/2]<<endl;
}