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>
#include<vector>
#include<queue>
#include<cstring>
#include<cmath>
#include<map>

#define LL  long long
#define MAXN    200010
#define P   1000000007
using namespace std;
int N;
int A[MAXN];
map<int, LL> dp[MAXN];

int main()
{
    
    cin>>N;

    for(int i=1; i<=N; i++) {
        cin>>A[i];
    }    

	dp[1][0] = 0; dp[1][1] = A[1];
	dp[2][0] = 0; dp[2][1] = max(A[1], A[2]);
	for(int i=3; i<=N; i++) {
		for(int j=i/2-1; j<=(i+1)/2; j++) {
			if(dp[i-1].count(j)) {
				dp[i][j] = max(dp[i][j], dp[i-1][j]);
			}
			if(dp[i-2].count(j-1)) {
				dp[i][j] = max(dp[i][j], dp[i-2][j-1]+A[i]);
			}
		}
	}

	cout << dp[N][N/2]<<endl;
    return 0;
}