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<algorithm>
using namespace std;
const int MAX = 200001;
int N;
bool last(int num,int n){
	if(num==N-1-n)return true;
	return false;
}
int main(){
	int A[MAX];
	long long sum[MAX][3],sum1=0,sum2=0,ans=0;
	bool visit[MAX][3]={0};
	cin>>N;
	for(int i=0;i<N;i++)cin>>A[i];
	if(N%2==0){
		for(int i=0;i<N;i+=2)sum1+=A[i];
		for(int i=1;i<N;i+=2)sum2+=A[i];
		cout<<max(sum1,sum2);
		return 0;
	}
	for(int i=0;i<N-1;i+=2)ans+=A[i];
	sum[0][2]=A[0];
	sum[1][1]=A[1];
	sum[2][0]=A[2];
	for(int i=0;i+2<N;i++){
		for(int j=0;j<3;j++){
			for(int k=0;k<=j;k++){
				if(i+2+k<N){
					if(visit[i+2+k][j-k])sum[i+2+k][j-k]=max(sum[i+2+k][j-k],sum[i][j]+A[i+2+k]);
					else sum[i+2+k][j-k]=sum[i][j]+A[i+2+k];
					visit[i+2+k][j-k]=true;
					if(last(i+2+k,j-k))ans=max(ans,sum[i+2+k][j-k]);
				}
			}
		}
	}
	cout<<ans;
	return 0;
}