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: Accepted)

/*
 * じょえチャンネル
 * 高評価・チャンネル登録よろしくおねがいします！
 * https://www.youtube.com/channel/UCRXsI3FL_kvaVL9zoolBfbQ
 */

#include <algorithm>
#include <bitset>
#include <cassert>
#include <cfloat>
#include <climits>
#include <cmath>
#include <complex>
#include <ctime>
#include <deque>
#include <fstream>
#include <functional>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <list>
#include <map>
#include <memory>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <string>
#include <string.h>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>

#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")




//here!!!
// define int long long !!!!!

#define int long long

// define int long long !!!!!


#define mod 1000000007ll
//constexpr int mod = 998244353ll;

constexpr double PI=3.141592653589793238462643383279;
//constexpr double eps = DBL_EPSILON;

typedef long long ll;

#ifdef int
#define inf (int)(3e18)
#else
#define inf (int)(5e8)
#endif

#define intt long long
#define itn long long
#define innt long long
#define P pair<long long,long long>

#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,n) for(int i=1;i<=n;i++)
#define rev_rep(i,n) for(int i=n-1;i>=0;i--)
#define REV_REP(i,n) for(int i=n;i>=1;i--)

#define ALL(v) v.begin(),v.end()

#define smallpriority_queue(T) priority_queue<T,vector<T>,greater<T>>


using namespace std;

//Library
//モッドパウ
inline int modpow(int x, int y, int m = mod) {
    int res = 1;
    while (y) {
        if (y & 1) {
            res *= x;
            res %= m;
        }
        x = x * x % m;
        y /= 2;
    }
    return res;
}

int mypow(int x, int y) {
    int res = 1;
    while (y) {
        if (y % 2) {
            res *= x;
        }
        x = x * x;
        y /= 2;
    }
    return res;
}
//is the number (x) a prime number?
//bool prime(int x) {
//    if (!x || x == 1) {
//        return false;
//    }
//    for (int i = 2; i * i <= x; i++) {
//        if (!(x % i)) {
//            return false;
//        }
//    }
//    return true;
//}
bool prime(int x) {
    if (!x || x == 1) {
        return false;
    }
    if(x == 2){
        return true;
    }
    if(!(x & 1)){
        return false;
    }
    for (int i = 3; i * i <= x; i+=2) {
        if (!(x % i)) {
            return false;
        }
    }
    return true;
}


//saidai-kouyakusuu
inline int gcd(int x, int y) {
    if (!y) {
        return x;
    }
    return gcd(y, x % y);
}


//number of keta
int keta(int x) {
    int ans = 0;
    while (x) {
        x /= 10;
        ans++;
    }
    return ans;
}

//number of 2shinsuu keta
int bitketa(int x) {
    int ans = 0;
    while (x) {
        x >>= 1;
        ans++;
    }
    return ans;
}

//sum of keta
int ketasum(int x) {
    int ans = 0;
    while (x) {
        ans += x % 10;
        x /= 10;
    }
    return ans;
}

inline int lcm(int x, int y) {
    int ans = x / gcd(x, y) * y;
    return ans;
}
int twobeki(int x) {
    int ans = 0;
    while (1) {
        if (!(x & 1)) {
            ans++;
            x >>= 1;
        }
        else {
            break;
        }
    }
    return ans;
}


template <class T, class U>
inline bool chmax(T& lhs, const U& rhs) {
    if (lhs < rhs) {
        lhs = rhs;
        return 1;
    }
    return 0;
}
template <class T, class U>
inline bool chmin(T& lhs, const U& rhs) {
    if (lhs > rhs) {
        lhs = rhs;
        return 1;
    }
    return 0;
}
void Yes(){
    cout<<"Yes"<<endl;
}
void No(){
    cout<<"No"<<endl;
}
void YES(){
    cout<<"YES"<<endl;
}
void NO(){
    cout<<"NO"<<endl;
}


#define fin(i) scanf("%lld",&i)
#define fout(i) printf("%lld",i)
#define fendl printf("\n")

int kai(int x, int y) {
    int res = 1;
    for (int i = x - y + 1; i <= x; i++) {
        res *= i; res %= mod;
    }
    return res;
}

int comb(int x, int y) {
    if (y > x)return 0;
    //    cout<<kai(x, y)<<' '<<modpow(kai(y, y), mod - 2)<<endl;
    return kai(x, y) * modpow(kai(y, y), mod - 2) % mod;
}
//Library-End

//入出力高速化時にはoff！！！！
//#define vecin(v) for(int i=0;i<v.size();i++)scanf("%lld",&v[i]);
//#define vecout(v) {if(v.size())printf("%lld",v[0]);for(int i=1;i<(int)v.size();i++)printf(" %lld",v[i]);printf("\n");}



template <class T>
class SegTree {
    int n;
    vector<T> node;
    T def;
    function<T(T,T)> operation;
    function<T(T,T)> update;
    
public:
    SegTree(unsigned int _n, T _def, function<T(T, T)> _operation,
            function<T(T, T)> _update)
    : def(_def), operation(_operation), update(_update) {
        n=1;
        while (n < _n) {
            n *= 2;
        }
        node = vector<T>(n * 2, def);
    }
    SegTree(vector<int>& initvec, function<T(T, T)> _operation,
            function<T(T, T)> _update)
    :  operation(_operation), update(_update) {
        n=1;
        while (n < initvec.size()) {
            n *= 2;
        }
        node = vector<T>(n * 2, def);
        for(int i=n;i<n+initvec.size();i++){
            node[i]=initvec[i-n];
        }
        for(int i=n-1;i>=1;i--){
            node[i]=operation(node[i*2],node[i*2+1]);
        }
    }
    void change(int i,T x){
        i+=n;
        node[i]=update(node[i],x);
        while (i>=1) {
            i>>=1;
            node[i]=operation(node[i*2],node[i*2+1]);
        }
    }
    T query(int l, int r){
        l+=n;
        r+=n;
        T rx=def,lx=def;
        while(l<r){
            if (l&1) {
                lx=operation(lx,node[l]);
                l++;
            }
            if (r&1) {
                r--;
                rx=operation(node[r],rx);
            }
            l>>=1; r>>=1;
        }
        return operation(lx,rx);
    }
    T operator [] (const int& x){
        return node[x+n];
    }
    void fill(T x) {
        std::fill(ALL(node), x);
    }
    void print() {
        rep(i, n)std::cout << operator[](i) << " ";
        std::cout << std::endl;
    }
    T queryForALL(){
        return node[1];
    }
};

#define R_MIN ([](long long a, long long b) { return min(a, b); })
#define R_MAX ([](long long a, long long b) { return max(a, b); })
#define R_SUM ([](long long a, long long b) { return a + b; })

#define NORMAL_UPDATE ([](long long a, long long b) { return b; })
#define ADD_UPDATE ([](long long a, long long b) { return a + b; })
#define MINUS_UPDATE ([](long long a, long long b) { return a - b; }

class Union_Find {
    vector<int> par;
    vector<int> ookisa;
    
public:
    Union_Find(int size) {
        par = vector<int>(size);
        ookisa=vector<int>(size);
        for (int i = 0; i < size; i++) {
            par[i] = i;
            ookisa[i]=1;
        }
    }
    
    int find(int x) {
        if (par[x] == x) {
            return x;
        }
        return par[x] = find(par[x]);
    }
    
    void unite(int x, int y) {
        x = find(x);
        y = find(y);
        if (x == y) {
            return;
        }
        if (ookisa[x] < ookisa[y]) {
            par[x] = y;
            ookisa[y]+=ookisa[x];
            ookisa[x]=0;
        }
        else {
            par[y] = x;
            ookisa[x]+=ookisa[y];
            ookisa[y]=0;
        }
    }
    int size(int i){
        i=find(i);
        return ookisa[i];
    }
    bool same(int x, int y){
        return find(x) == find(y);
    }
    
};

class BIT {
    vector<int> data;
    int size=0;
public:
    BIT(int _size){
        data=vector<int>(_size+1);
        size=_size;
    }
    void add(int i,int x){
        while (i<=size) {
            data[i]+=x;
            i += i & -i;
        }
    }
    int sum(int i){
        assert(i<=size);
        int s=0;
        while(i>0){
            s+=data[i];
            i -= i & -i;
        }
        return s;
    }
    int lower_bound(int x){
        if(x<=0){
            return 0;
        }else{
            int i=0;int r=1;
            while(r<size) r=r<<1;
            for(int len=r;len>0;len=len>>1) {
                if(i+len<size && data[i+len]<x){
                    x-=data[i+len];
                    i+=len;
                }
            }
            return i+1;
        }
    }
};

//Union-Find-End

int perm[2000005];
void init_perm() {
    perm[0] = 1;
    REP(i, 2000004) {
        perm[i] = perm[i - 1] * i % mod;
    }
}

int nCk(int x, int y) {
    if (y>x) {
        return 0;
    }
    if (x<0) {
        return 0;
    }
    return perm[x] * modpow(perm[x - y], mod - 2) % mod * modpow(perm[y], mod - 2) % mod;
}

struct Dot {
    double x;
    double y;
    Dot(double _x=0.0, double _y=0.0){
        x=_x;
        y=_y;
    }
};

struct Dot_i {
    int x;
    int y;
    Dot_i(int _x=0, int _y=0){
        x=_x;
        y=_y;
    }
};

double kyori(pair<int, int> f, pair<int, int> s) {
    double ans = 0;
    double t = fabs(f.first - s.first);
    double y = fabs(f.second - s.second);
    ans = sqrt(t * t + y * y);
    return ans;
}

double kyori(Dot f, Dot s) {
    double ans = 0;
    double t = fabs(f.x - s.x);
    double y = fabs(f.y - s.y);
    ans = sqrt(t * t + y * y);
    return ans;
}

inline string stringmax(string& x,string& y){
    if (x.size()>y.size()) {
        return x;
    }
    if (x.size()<y.size()) {
        return y;
    }
    rep(i,x.size()){
        if (x[i]>y[i]) {
            return x;
        }
        if (x[i]<y[i]) {
            return y;
        }
    }
    return x;
}

//vector<int>  RollingHash(string &s, string& t){
//    vector<int> ans;
//    __int128 h=0,hamod=0,ki=0,kim=0,hikaku=0,one=0;
//    one=1;
//    ki=1000000007ll;
//    hamod=(one<<61)-1;
//    kim=1;
//    rep(i,t.size()){
//        hikaku*=ki;
//        h*=ki;
//        kim*=ki;
//        hikaku+=t[i];
//        h+=s[i];
//        hikaku%=hamod;
//        h%=hamod;
//        kim%=hamod;
//    }
//    rep(i,(int)s.size()-(int)t.size()+1){
//        if (h==hikaku) {
//            ans.emplace_back(i);
//        }
//        h*=ki;
//        h%=hamod;
//        h+=s[i+(int)t.size()];
//        h%=hamod;
//        h+=hamod;
//        h-=s[i]*kim%hamod;
//        h%=hamod;
//    }
//    return ans;
//}
struct edge {
    int to;
    int length;
    edge(int _to, int _length){
        to=_to;
        length=_length;
    }
};
vector<int> djkstra(vector<vector<edge>> &road,int start){
    vector<int> kyo(road.size(),inf);
    smallpriority_queue(P) q;
    q.push({0,start});
    kyo[start]=0;
    while (q.size()) {
        int x=q.top().second;
        itn now=q.top().first;
        q.pop();
        if (kyo[x]<now) {
            continue;
        }
        for(auto&i:road[x]){
            if (kyo[i.to]>now+i.length) {
                kyo[i.to]=now+i.length;
                q.push({kyo[i.to],i.to});
            }
        }
    }
    return kyo;
}

template <class T>
void change_to_unique(vector<T> &v){
    std::sort(ALL(v));
    auto k=unique(ALL(v));
    if(k!=v.end())v.erase(k,v.end());
}

double kodo_dosuu(double r){
    return 180.0*r/(double)PI;
}

double dosuu_kodo(double r){
    return r*(double)PI/180.0;
}

double kakudo(Dot a,Dot b1,Dot b2){
    double size1=kyori(a,b1),size2=kyori(a,b2);
    double nai=(b1.x-a.x)*(b2.x-a.x)+(b1.y-a.y)*(b2.y-a.y);
    nai/=size1*size2;
    return kodo_dosuu(acos(nai));
}

struct fraction {
    int shi;
    int bo;
    fraction(int bunshi,int bunbo):bo(bunbo),shi(bunshi){
        if(bunbo&&bunshi){
            int g=gcd(bo, shi);
            bo/=g;
            shi/=g;
        }
    };
    explicit inline operator double()const { return (double)shi/(double)bo; }
    explicit inline operator long double()const { return (long double)shi/(long double)bo; }
};
bool operator < (const fraction& b1, const fraction& b2){
    return b1.shi*b2.bo < b2.shi*b1.bo;
}
bool operator > (const fraction& b1, const fraction& b2){
    return b1.shi*b2.bo > b2.shi*b1.bo;
}

template <class T>
void vecout(vector<T> &v){
    if (v.size()) {
        cout<<v[0];
    }
    REP(i,(int)v.size()-1){
        cout<<' '<<v[i];
    }
    cout<<endl;
}


#define endl "\n" //interactive の時に注意！！！
#define Endl "\n" //interactive の時に注意！！！
#define cinn cin
#define printd cout<<fixed<<setprecision(10)
#define rrep(i,f,s) for(int i=f;i<s;i++)
#define RREP(i,f,s) for(int i=f;i<=s;i++)
#define REV_RREP(i,f,s) for(int i=s;i>=f;i--)
#define hen_rep(i,l,step) for(int i=0;i<l;i+=step)
#define HEN_REP(i,l,step) for(int i=1;i<=l;i+=step)


int n,ans;
itn s[100004];
vector<P> now[100004];
signed main() {
    ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    cin>>n;
    rep(i,n){
        cin>>s[i];
    }
    REP(i,n-1){
        int cnt=0;
        for(int j=n-i-1;j>=i+1;j-=i){
            cnt+=s[j];
            now[j].emplace_back(i,cnt);
        }
    }
    ans=0;
    REP(i,n-2){
        for(auto&j:now[i]){
            int ka=j.second;
            int k=j.first;
            if (!((i-j.first)%j.first)) {
                if (n-1>i+(i-j.first)) {
                    continue;
                }
            }
            while (k+(i-j.first)<n-1) {
                ka+=s[k];
                k+=j.first;
            }
            chmax(ans, ka);
        }
    }
    cout<<ans<<endl;
}