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>
// clang-format off
using namespace std; using ll = int64_t; using ull = uint64_t; const ll INF = 9e18;
void print() { cout << endl; }
template<typename Head,typename... Tail> void print(Head head,Tail... tail){cout<<head;if(sizeof...(Tail)>0)cout<<" ";print(tail...);}
void print0() {}
template<typename Head,typename... Tail> void print0(Head head,Tail... tail){cout<<head;print0(tail...);}
// clang-format on

ll dpval(vector<vector<ll>> &dp, ll i, ll j) {
    if (j >= 0) {
        return dp[i][j];
    }
    return -INF;
}
int main() {
    ll N;
    cin >> N;
    vector<ll> A(N);
    for (ll i = 0; i < N; i++) {
        cin >> A[i];
    }

    if (N % 2 == 0) {
        vector<vector<ll>> dp(2, vector<ll>(N, -INF));
        dp[0][0] = A[0];
        dp[1][1] = A[1];
        for (ll i = 2; i < N; i++) {
            dp[0][i] = A[i] + dpval(dp, 0, i - 2);
            dp[1][i] = A[i] + max(dpval(dp, 1, i - 2), dpval(dp, 0, i - 3));
        }
        ll result = max(dpval(dp, 0, N - 2), dpval(dp, 1, N - 1));
        print(result);
    } else {
        vector<vector<ll>> dp(3, vector<ll>(N, -INF));
        dp[0][0] = A[0];
        dp[1][1] = A[1];
        dp[2][2] = A[2];
        dp[0][2] = A[0] + A[2];
        for (ll i = 3; i < N; i++) {
            dp[0][i] = A[i] + dpval(dp, 0, i - 2);
            dp[1][i] = A[i] + max(dpval(dp, 1, i - 2), dpval(dp, 0, i - 3));
            dp[2][i] = A[i] + max(max(dpval(dp, 2, i - 2), dpval(dp, 1, i - 3)), dpval(dp, 0, i - 4));
        }
        ll result = max(max(dpval(dp, 0, N - 3), dpval(dp, 1, N - 2)), dpval(dp, 2, N - 1));
        print(result);
    }
}