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 Rust (1.15.1) (judge verdict: Wrong Answer)

#[allow(unused_macros)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}
 
#[allow(unused_macros)]
macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};
 
    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}
 
#[allow(unused_macros)]
macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };
 
    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };
 
    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };
 
    ($next:expr, bytes) => {
        read_value!($next, String).into_bytes()
    };
 
    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };
 
    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}
 
use std::cmp::Ordering;
use std::cmp;
use std::cmp::min;
use std::collections::BTreeMap;
use std::process;
use std::cmp::Ord;
use std::collections::HashMap;
use std::collections::HashSet;
use std::collections::VecDeque;
use std::collections::BTreeSet;
use std::mem;
use std::collections::BinaryHeap;
/// Equivalent to std::lowerbound and std::upperbound in c++
pub trait BinarySearch<T> {
    fn lower_bound(&self, &T) -> usize;
    fn upper_bound(&self, &T) -> usize;
}
 
impl<T: Ord> BinarySearch<T> for [T] {
    fn lower_bound(&self, x: &T) -> usize {
        let mut low = 0;
        let mut high = self.len();
 
        while low != high {
            let mid = (low + high) / 2;
            match self[mid].cmp(x) {
                Ordering::Less => {
                    low = mid + 1;
                }
                Ordering::Equal | Ordering::Greater => {
                    high = mid;
                }
            }
        }
        low
    }
 
    fn upper_bound(&self, x: &T) -> usize {
        let mut low = 0;
        let mut high = self.len();
 
        while low != high {
            let mid = (low + high) / 2;
            match self[mid].cmp(x) {
                Ordering::Less | Ordering::Equal => {
                    low = mid + 1;
                }
                Ordering::Greater => {
                    high = mid;
                }
            }
        }
        low
    }
}
 
fn matmul(A:&Vec<Vec<i64>>, B:&Vec<Vec<i64>>) -> Vec<Vec<i64>>{
    let mut C = vec![vec![0;B[0].len()];A.len()];
    for i in 0..A.len(){
        for k in 0..B.len(){
            for j in 0..B[0].len(){
                C[i][j] += A[i][k]*B[k][j];
                C[i][j] %= MOD;
            }
        }
    }
    return C;
}
fn matpow(A:&mut Vec<Vec<i64>>, n:usize) -> Vec<Vec<i64>>{
    let mut B = vec![vec![0;A.len()];A.len()];
    for i in 0..A.len(){
        B[i][i] = 1;
    }
    let mut n = n;
    let mut tmp = A.clone();
    while(n>0){
        if n&1 == 1{B = matmul(&B, &tmp);}
        tmp = matmul(&tmp, &tmp);
        n>>=1;
    }
    return B;
}

 
 
 

/*
subset enumeration
let mut set;
let mut subset = set;
        for sm2 in 0..1<<n{
            if subset==0{
                break;
            }

            hogehoge


            subset = (subset-1)&set;
        }
*/
/*
全方位木
    fn dfs(s:usize,bef:usize,m:i64, g:&Vec<Vec<usize>>, dp:&mut Vec<Vec<i64>>)->i64{
        let mut res = 1;
        for i in 0..g[s].len(){
            let e = g[s][i];
            if e != bef{
                dp[s][i] = dfs(e, s, m, g, dp);
                res *= (1 + dp[s][i]);
                res %= m;
            }
        }
        return res;
    }
    let tmp = dfs(0, n, m, &g, &mut dp);
    fn dfs2(s:usize, bef:usize, p:i64, m:i64, g:&Vec<Vec<usize>>, dp:&mut Vec<Vec<i64>>){
        for i in 0..g[s].len(){
            let e = g[s][i];
            if e == bef{dp[s][i] = p;}
        }
        let mut L = vec![0;g[s].len()];
        let mut R = vec![0;g[s].len()];

        for i in 0..g[s].len(){
            L[i] = {
                let mut tmp = 1+dp[s][i];
                tmp %= m;
                if i != 0{
                    tmp *= L[i-1];
                }
                tmp%m
            }
        }

        for i in (0..g[s].len()).rev(){
            if i+1<g[s].len(){
                R[i] = (R[i+1]*((1+dp[s][i])%m))%m;
            }
            else{
                R[i] = (1+dp[s][i])%m;
            }

            if g[s][i] == bef{
                continue;
            }  
            let mut tmp = 1;
            if i != 0{
                tmp *= L[i-1];
                tmp %= m;
            }
            if i+1 != g[s].len(){
                tmp *= R[i+1];
                tmp %= m;
            }
            dfs2(g[s][i], s, tmp, m, g, dp);
        }
    }

    */
fn mpow(x:i64, n:i64) -> i64{
    let mut ans = 1;
    let mut n = n;
    let mut x = x;
    while(n != 0){
        if (n&1 == 1){ans = ans*x%MOD;}
        x = x*x%MOD;
        n = n>>1;
    }
    ans
}
fn comb(a:i64, b:i64, fac:&Vec<i64>, ifac:&Vec<i64>) -> i64{
    if a == 0 && b == 0{return 1;}
    if a<b || a<0 {return 0;}
    let mut tmp = ifac[(a-b) as usize]*ifac[b as usize] %MOD;
    return tmp*fac[a as usize] %MOD;
}
fn divisor(n:usize) -> Vec<usize>{
    let mut res:Vec<usize> = Vec::new(); 
    for i in 1..n+1{
        if i*i>n{break;}
        if n%i == 0{
            res.push(i);
            if i != n/i{
                res.push(n/i);    
            }    
        }    
    }
    res
}
struct UnionFind{
    par:Vec<usize>,
    rank:Vec<usize>,
    size:Vec<usize>,
}
impl UnionFind{
    fn init(n:usize) -> UnionFind{
        let mut par = vec![0;n];
        for i in 0..n{
            par[i] = i;
        }
        UnionFind{
            par:par,
            rank:vec![0;n],
            size:vec![1;n],
        }
    }
    fn find(&mut self, x:usize) ->usize{
        if(self.par[x] == x){
            x
        }
        else{
            let p = self.par[x];
            let res = self.find(p);
            self.par[x] = res;
            res
        }
    }
    fn same(&mut self, a:usize, b:usize)->bool{
        self.find(a) == self.find(b)
    }
    fn unite(&mut self, a:usize, b:usize){
        let x = self.find(a);
        let y = self.find(b);
        if x != y{
            if (self.rank[x] < self.rank[y]){
                self.par[x] = y;
                self.size[y] += self.size[x];
            }
            else{
                self.par[y] = x;
                self.size[x] += self.size[y];
                if(self.rank[x] == self.rank[y]){ self.rank[x]+=1;}
            }
        }
    }
    fn check_size(&mut self, a:usize) -> usize{
        let x = self.find(a);
        let s = self.size[x];
        s
    }
}
pub static MOD:i64 = 1000000007;
const INF: i64 = 1 << 60;
fn main(){
    input!{
        n:usize,
        s:[i64;n],
    }
    let mut res = 0;
    for D in 1..n-1{
        let mut l = 0;
        let mut r = n-1;
        let mut rs = 0;
        let mut ls = 0;
        let mut used = vec![false;n];
        while(true){
            if l>=r || l + D >=n-1 || r<D || used[l] || used[r]{
                break;
            }
            rs += s[r];
            ls += s[l];
            res = cmp::max(res, rs + ls);
            used[r] = true;
            used[l] = true;
            l+=D;
            r-=D;


        }
    }
    println!("{}", res);
}
pub mod fenwick_tree {
    use std::ops::{AddAssign, Sub};
    /// `FenwickTree` is a data structure that can efficiently update elements
    /// and calculate prefix sums in a table of numbers.
    /// [https://en.wikipedia.org/wiki/Fenwick_tree](https://en.wikipedia.org/wiki/Fenwick_tree)
    pub struct FenwickTree<T> {
        n: usize,
        data: Vec<T>,
        init: T,
    }
 
    impl<T: Copy + AddAssign + Sub<Output = T>> FenwickTree<T> {
        /// Constructs a new `FenwickTree`. The size of `FenwickTree` should be specified by `size`.
        pub fn new(size: usize, init: T) -> FenwickTree<T> {
            FenwickTree {
                n: size + 1,
                data: vec![init; size + 1],
                init: init,
            }
        }
 
        pub fn add(&mut self, k: usize, value: T) {
            let mut x = k;
            while x < self.n {
                self.data[x] += value;
                x |= x + 1;
            }
        }
 
        /// Returns a sum of range `[l, r)`
        pub fn sum(&self, l: usize, r: usize) -> T {
            self.sum_one(r) - self.sum_one(l)
        }
 
        /// Returns a sum of range `[0, k)`
        pub fn sum_one(&self, k: usize) -> T {
            assert!(k < self.n, "k={} n={}", k, self.n);
 
            let mut result = self.init;
            let mut x = k as i32 - 1;
            while x >= 0 {
                result += self.data[x as usize];
                x = (x & (x + 1)) - 1;
            }
 
            result
        }
    }
}