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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll INF=1e18;

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N; cin >> N;
    vector<ll> A(N);
    for (int i=0;i<N;++i) cin >> A[i];
    vector<vector<ll>> dp(N+1,vector<ll>(4,-INF));
    dp[0][(N&1?0:2)]=0;
    for (int i=0;i<N;++i){
        dp[i+1][1]=max(dp[i+1][1],dp[i][0]);
        dp[(min(N,i+2))][0]=max(dp[(min(N,i+2))][0],dp[i][0]+A[i]);
        dp[(min(N,i+2))][2]=max(dp[(min(N,i+2))][2],dp[i][1]+A[i]);
        dp[(min(N,i+2))][2]=max(dp[(min(N,i+2))][2],dp[i][2]+A[i]);
        dp[(min(N,i+2))][3]=max(dp[(min(N,i+2))][3],dp[i][1]+A[i+1]);
        dp[(min(N,i+2))][3]=max(dp[(min(N,i+2))][3],dp[i][2]+A[i+1]);
        dp[(min(N,i+2))][3]=max(dp[(min(N,i+2))][3],dp[i][3]+A[i+1]);
    }
    cout << max({dp[N][1],dp[N][2],dp[N][3]}) << '\n';
}