Problem:
Score : 600 points

Problem Statement

There is an infinitely large pond, which we consider as a number line.
In this pond, there are N lotuses floating at coordinates 0, 1, 2, ..., N-2 and N-1.
On the lotus at coordinate i, an integer s_i is written.

You are standing on the lotus at coordinate 0. You will play a game that proceeds as follows:

1. Choose positive integers A and B. Your score is initially 0.

2. Let x be your current coordinate, and y = x+A. The lotus at coordinate x disappears, and you move to coordinate y.

If y = N-1, the game ends.

If y \neq N-1 and there is a lotus floating at coordinate y, your score increases by s_y.

If y \neq N-1 and there is no lotus floating at coordinate y, you drown. Your score decreases by 10^{100} points, and the game ends.

3. Let x be your current coordinate, and y = x-B. The lotus at coordinate x disappears, and you move to coordinate y.

If y = N-1, the game ends.

If y \neq N-1 and there is a lotus floating at coordinate y, your score increases by s_y.

If y \neq N-1 and there is no lotus floating at coordinate y, you drown. Your score decreases by 10^{100} points, and the game ends.

4. Go back to step 2.

You want to end the game with as high a score as possible.
What is the score obtained by the optimal choice of A and B?

Constraints

3 \leq N \leq 10^5

-10^9 \leq s_i \leq 10^9

s_0=s_{N-1}=0

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
s_0 s_1 ...... s_{N-1}

Output

Print the score obtained by the optimal choice of A and B.

Sample Input 1

5
0 2 5 1 0

Sample Output 1

3

If you choose A = 3 and B = 2, the game proceeds as follows:

Move to coordinate 0 + 3 = 3. Your score increases by s_3 = 1.

Move to coordinate 3 - 2 = 1. Your score increases by s_1 = 2.

Move to coordinate 1 + 3 = 4. The game ends with a score of 3.

There is no way to end the game with a score of 4 or higher, so the answer is 3. Note that you cannot land the lotus at coordinate 2 without drowning later.

Sample Input 2

6
0 10 -7 -4 -13 0

Sample Output 2

0

The optimal strategy here is to land the final lotus immediately by choosing A = 5 (the value of B does not matter).

Sample Input 3

11
0 -4 0 -99 31 14 -15 -39 43 18 0

Sample Output 3

59

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <queue>
#include <string>
#include <map>
#include <set>
#include <stack>
#include <tuple>
#include <deque>
#include <numeric>
#include <bitset>
#include <iomanip>
#include <cassert>
#include <chrono>
#include <random>
#include <limits>
#include <iterator>
#include <functional>
#include <sstream>
#include <complex>
using namespace std;

using ll = long long;
using ull = uint64_t;
using P = pair<int, int>;
constexpr double EPS = 1e-9;
constexpr int INF = 1001001001;
constexpr int mod = 1000000007;
// constexpr int mod = 998244353;

template<class T>
inline bool chmax(T& x, T y){
    if(x < y){
        x = y;
        return true;
    }
    return false;
}
template<class T>
inline bool chmin(T& x, T y){
    if(x > y){
        x = y;
        return true;
    }
    return false;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n;
    cin >> n;
    vector<int> s(n);
    ll ans = 0;
    for(int i = 0; i < n; ++i){
        cin >> s[i];
        ans += s[i];
    }
    if(n % 2 != 0)  ans -= s[n / 2];

    for(int k = 2; k < n / 2; ++k){
        vector<ll> LeftSum(1, 0), RightSum(1, 0);
        if((n - 1) % k != 0){
            for(int i = n - k - 1; i > k; i -= k){
                RightSum.emplace_back(RightSum.back() + s[i]);
            }
            int sz = RightSum.size();
            for(int i = k; i < k * sz; i += k){
                LeftSum.emplace_back(LeftSum.back() + s[i]);
            }
            for(int i = 0; i < sz; ++i){
                chmax(ans, LeftSum[i] + RightSum[i]);
            }
        }
        else{
            for(int i = k; i < n / 2; i += k){
                LeftSum.emplace_back(LeftSum.back() + s[i]);
            }
            for(int i = n - k - 1; i >= n / 2; i -= k){
                RightSum.emplace_back(RightSum.back() + s[i]);
            }
            int sz = LeftSum.size();
            if(sz != (int)RightSum.size()){
                for(int i = 0; i < sz; ++i){
                    chmax(ans, LeftSum[i] + RightSum[i + 1]);
                }
            }
            else{
                for(int i = 0; i < sz; ++i){
                    chmax(ans, LeftSum[i] + RightSum[i]);
                }
            }
        }
    }

    cout << ans << endl;
}