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 <bits/stdc++.h>
#define err(args...) {}
#ifdef DEBUG
#include "_debug.cpp"
#endif
using namespace std;
using ll = long long;
using ld = long double;
template <typename T> using lim = numeric_limits<T>;
template <typename T> istream& operator>>(istream& is, vector<T>& a) { for(T& x : a) { is >> x; } return is; }
template <typename X, typename Y> istream& operator>>(istream& is, pair<X, Y>& p) { return is >> p.first >> p.second; }
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    ll n;
    cin >> n;
    ll rt_n = int(sqrt(n));
    vector<ll> s(n);
    cin >> s;
    vector<vector<vector<ll>>> modsum(rt_n);
    for(int c = 1; c < rt_n; c++) {
        modsum[c] = vector<vector<ll>>(c);
        for(int i = 0; i < n; i++) {
            modsum[c][i % c].push_back(s[i]);
        }
        for(int j = 0; j < c; j++) {
            partial_sum(modsum[c][j].begin(), modsum[c][j].end(), modsum[c][j].begin());
        }
    }
    auto rsq = [](vector<ll>& p, int L, int R) {
        return p[R - 1] - (L ? p[L - 1] : 0);
    };
    ll ans = 0;
    for(ll c = 1; c <= n; c++) {
        for(ll a = (n - 1) % c; a < n - 1; a += c) {
            ll b = a - c;
            ll steps = a / gcd(a, b);
            if(b > 0 and b < a and a + c * (steps - 1) >= n - 1) {
                ll cur = 0;
                if(c >= rt_n) {
                    for(int i = a, step = 0; i < n and step < (n - 1 - a) / c; i += c, step++) {
                        cur += s[i];
                    }
                    for(int i = a - b, step = 0; i < n and step < (n - 1 - a) / c; i += c, step++) {
                        cur += s[i];
                    }
                } else {
                    int L1 = a / c, L2 = (a - b) / c;
                    cur += rsq(modsum[c][a % c], L1, L1 + (n - 1 - a) / c);
                    cur += rsq(modsum[c][(a - b) % c], (a - b) / c, L2 + (n - 1 - a) / c);
                }
                ans = max(ans, cur);
            }
        }
    }
    cout << ans << endl;
    return 0;
}