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++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double Double;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)

// chmax, chmin
template<class T>
inline bool chmax(T &a, T b) {
    if (a < b) {
        a = b;
        return 1;
    }
    return 0;
}

template<class T>
inline bool chmin(T &a, T b) {
    if (a > b) {
        a = b;
        return 1;
    }
    return 0;
}


long long infinite(long long A, long long B, long long N) {
    return __gcd<ll>(A, B) == A - B && N - 1 - B > A;
}


ll sum[20][20][100010];


long long simulate(long long A, long long B, vector<ll> &s) {
    if (infinite(A, B, s.size())) return -1e18;

    ll x = A;
    ll sc = s[A];

    int k = (s.size() - 1 - B) / (A - B);
    assert((s.size() - 1 - B) % (A - B) == 0);

    if (A - B < 20) {
        sc += sum[A - B][0][(k - 1) * (A - B)];
        sc += sum[A - B][A][(k - 1) * (A - B) + A] - s[A];
    } else {
        for (int i = 0; i <= k - 1; i++) {
            sc += s[i * (A - B)];
        }
        for (int i = 1; i <= k - 1; i++) {
            sc += s[i * (A - B) + A];
        }
    }
    return sc;
}

vector<long long> get_divisors(long long n) {
    vector<long long> ret;
    for (long long i = 1; i * i <= n; i++) {
        if (n % i == 0) {
            ret.push_back(n / i);
            if (i != n / i)ret.push_back(i);
        }
    }
    sort(ret.begin(), ret.end());
    return ret;
}


void solve(long long N, std::vector<long long> S) {
    //N = 100000;
    //S.resize(N, 1);


    for (int d = 1; d < 20; d++) {
        for (int s = 0; s < 20; s++) {
            for (int i = s; i < N; i += d) {
                sum[d][s][i] += S[i];
            }
            for (int i = 1; i < N; i++) {
                sum[d][s][i] += sum[d][s][i - 1];
            }
        }
    }
    //N = 10000;
    //s.resize(N);
    // odd
    long long ans = 0;
    for (int B = 1; B <= N; B++) {
        if (N - 1 - B < 0) continue;
        auto ds = get_divisors(N - 1 - B);
        for (auto d : ds) {
            auto A = d + B;
            auto r = simulate(A, B, S);
            //assert((r == -44444) == infinite(A, B, N));
            //cout << A << " " << B << " " << r << " " << infinite(A, B, N) << endl;
            ans = max(ans, r);
        }
    }
    cout << ans << endl;
    cerr << "[]" << endl;
}

int main() {
    long long N;
    scanf("%lld", &N);
    std::vector<long long> s(N - 1 - 0 + 1);
    for (int i = 0; i < N - 1 - 0 + 1; i++) {
        scanf("%lld", &s[i]);
    }
    solve(N, std::move(s));
    return 0;
}