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<bits/stdc++.h>
using namespace std;

using i32 = int_fast32_t;
using i64 = int_fast64_t;
using p32 = pair<i32, i32>;

i32 n;
vector<i32> a;
map<p32, i64> dp;

i32 dfs(i32 index, i32 use) {
    if (use == n / 2) return 0;
    if (index / 2 < use) return -INT32_MAX;
    if ((n - index + 1) / 2 < n / 2 - use) return -INT32_MAX;
    if (dp.find({index, use}) != dp.end()) return dp[{index, use}];
    return dp[{index, use}] = max(a[index] + dfs(index + 2, use + 1), dfs(index + 1, use));
}

int main() {
    cin >> n;
    a.assign(n, 0);
    for (auto& e : a) cin >> e;
    cout << dfs(0, 0) << endl;
}