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>
using namespace std;
typedef long long int64;
const int maxn = 2e5 + 10;//
map<int, map<int, int64>> dp;
int main(){
    int64 n; cin >> n;
    vector<int64> A(n);
    for(int i = 0; i < n; i++){
        cin >> A[i];
    }
    int64 ans = 0;
    dp[0][0] = 0; dp[0][1] = A[0];
    for(int i = 1; i < n; i++){
        for(int taken = 1; taken <= ((i + 1) / 2); taken++){
            dp[i][taken] = max(A[i] + ((i >= 2) ? dp[i - 2][taken - 1] : 0LL),
                               dp[i - 1][taken]);
            //cout << "i = " << i << " taken = " << taken << " = " << dp[i][taken] << endl;
        }
    }
    cout << dp[n - 1][n / 2] << endl;
    return 0;
}