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: Accepted)

#include <iostream>
#include <algorithm>

#define INF 1000000000000000

using namespace std;

int main(){
    int N;
    cin >> N;
    long A[N];
    for(int i=0; i<N; i++) cin >> A[i];
    long dp[N][3][2];

    // initialize
    for(int j=0; j<3; j++){
        for(int s=0; s<2; s++) dp[0][j][s] = -INF;
    }
    dp[0][0][0] = 0;
    dp[0][1][1] = A[0];

    long base[N];
    base[0] = 0;

    for(int i=1; i<N; i++){

        base[i] = N / 2 - (N - i) / 2;
        for(int j=base[i]; j<=(i+2)/2; j++){
            if(j  <= (i+1)/2) dp[i][j - base[i]][0] = max(dp[i-1][j - base[i-1]][0], dp[i-1][j - base[i-1]][1]);
            else dp[i][j - base[i]][0] = -INF;
            if(j-1 <= (i+1)/2) dp[i][j - base[i]][1] = dp[i-1][j-1-base[i-1]][0] + A[i];
            else dp[i][j - base[i]][1] = -INF;
        }
    }
    cout << max(dp[N-1][N/2 - base[N-1]][0], dp[N-1][N/2 - base[N-1]][1]) << endl;
    return 0;
}