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: Time Limit Exceeded)

#ifdef LOCAL
#define _GLIBCXX_DEBUG
#endif
#include<bits/stdc++.h>
using namespace std;
#define rep(i,s,t) for(ll i = (ll)(s); i < (ll)(t); i++)
#define rrep(i,s,t) for(ll i = (ll)(s-1);(ll)(t) <= i; i--)
#define all(x) (x).begin(), (x).end()
typedef long long ll;
typedef long double ld;
typedef pair<ll,ll> Pll;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<vvl> vvvl;
constexpr ll INF = numeric_limits<ll>::max()/4;
constexpr ll n_max = 2e5+10;
#define int ll

template <typename A, typename B>
string to_string(pair<A, B> p);
string to_string(const string &s) {return '"' + s + '"';}
string to_string(const char *c) {return to_string((string) c);}
string to_string(bool b) {return (b ? "true" : "false");}
template <size_t N>
string to_string(bitset<N> v){
    string res = "";
    for(size_t i = 0; i < N; i++) res += static_cast<char>('0' + v[i]);
    return res;
}
template <typename A>
string to_string(A v) {
    bool first = true;
    string res = "{";
    for(const auto &x : v) {
        if(!first) res += ", ";
        first = false;
        res += to_string(x);
    }
    res += "}";
    return res;
}
template <typename A, typename B>
string to_string(pair<A, B> p){return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";}

void debug_out() {cerr << endl;}
template<typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
    cerr << " " << to_string(H);
    debug_out(T...);
}

#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endif

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

signed main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    ll n; cin >> n;
    vector<ll> s(n);
    for(int i = 0; i < n; i++) cin >> s[i];
    ll ans = 0;
    // bitset<n_max> used(0);
    vector<ll> used(n);
    rep(c,1,n){
        ll lp = 0, rp = n-1;
        ll temp = 0;
        fill(all(used), 0);
        for(; lp < n; lp += c){
            if(lp == rp)break;
            if(rp < c)break;
            if(used[lp])break;
            temp += s[lp];
            temp += s[rp];
            chmax(ans, temp);
            debug(c, lp, rp, temp);
            rp -= c;
            used[lp] = used[rp] = 1;
        }
    }

    cout << ans << "\n";
}