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 (1.42.0) (judge verdict: Wrong Answer)

pub trait IterScan: Sized {
    type Output;
    fn scan<'a, I: Iterator<Item = &'a str>>(iter: &mut I) -> Option<Self::Output>;
}
pub trait MarkedIterScan: Sized {
    type Output;
    fn mscan<'a, I: Iterator<Item = &'a str>>(self, iter: &mut I) -> Option<Self::Output>;
}
#[derive(Debug)]
pub struct Scanner<'a> {
    iter: std::str::SplitAsciiWhitespace<'a>,
}
impl<'a> Scanner<'a> {
    #[inline]
    pub fn new(s: &'a str) -> Self {
        let iter = s.split_ascii_whitespace();
        Self { iter }
    }
    #[inline]
    pub fn scan<T: IterScan>(&mut self) -> <T as IterScan>::Output {
        T::scan(&mut self.iter).unwrap()
    }
    #[inline]
    pub fn mscan<T: MarkedIterScan>(&mut self, marker: T) -> <T as MarkedIterScan>::Output {
        marker.mscan(&mut self.iter).unwrap()
    }
    #[inline]
    pub fn scan_vec<T: IterScan>(&mut self, size: usize) -> Vec<<T as IterScan>::Output> {
        (0..size)
            .map(|_| T::scan(&mut self.iter).unwrap())
            .collect()
    }
    #[inline]
    pub fn scan_chars(&mut self) -> Vec<char> {
        self.iter.next().unwrap().chars().collect::<Vec<char>>()
    }
}
mod scanner_impls {
    use super::*;
    macro_rules !iter_scan_impls {($($t :ty ) *) =>{$(impl IterScan for $t {type Output =Self ;#[inline ] fn scan <'a ,I :Iterator <Item =&'a str >>(iter :&mut I ) ->Option <Self >{iter .next () ?.parse ::<$t >() .ok () } } ) *} ;}
    iter_scan_impls !(char u8 u16 u32 u64 usize i8 i16 i32 i64 isize f32 f64 u128 i128 String ) ;
    macro_rules !iter_scan_tuple_impl {($($T :ident ) *) =>{impl <$($T :IterScan ) ,*>IterScan for ($($T ,) *) {type Output =($(<$T as IterScan >::Output ,) *) ;#[inline ] fn scan <'a ,It :Iterator <Item =&'a str >>(_iter :&mut It ) ->Option <Self ::Output >{Some (($($T ::scan (_iter ) ?,) *) ) } } } ;}
    iter_scan_tuple_impl!();
    iter_scan_tuple_impl!(A);
    iter_scan_tuple_impl !(A B ) ;
    iter_scan_tuple_impl !(A B C ) ;
    iter_scan_tuple_impl !(A B C D ) ;
    iter_scan_tuple_impl !(A B C D E ) ;
    iter_scan_tuple_impl !(A B C D E F ) ;
    iter_scan_tuple_impl !(A B C D E F G ) ;
    iter_scan_tuple_impl !(A B C D E F G H ) ;
    iter_scan_tuple_impl !(A B C D E F G H I ) ;
    iter_scan_tuple_impl !(A B C D E F G H I J ) ;
    iter_scan_tuple_impl !(A B C D E F G H I J K ) ;
    pub struct ScannerIter<'a, 'b, T> {
        inner: &'b mut Scanner<'a>,
        _marker: std::marker::PhantomData<fn() -> T>,
    }
    impl<'a, 'b, T: IterScan> Iterator for ScannerIter<'a, 'b, T> {
        type Item = <T as IterScan>::Output;
        fn next(&mut self) -> Option<Self::Item> {
            T::scan(&mut self.inner.iter)
        }
    }
    impl<'a> Scanner<'a> {
        #[inline]
        pub fn iter<'b, T: IterScan>(&'b mut self) -> ScannerIter<'a, 'b, T> {
            ScannerIter {
                inner: self,
                _marker: std::marker::PhantomData,
            }
        }
    }
}
mod marker {
    use super::*;
    pub struct Usize1;
    impl IterScan for Usize1 {
        type Output = usize;
        #[inline]
        fn scan<'a, I: Iterator<Item = &'a str>>(iter: &mut I) -> Option<Self::Output> {
            usize::scan(iter).map(|x| x.wrapping_sub(1))
        }
    }
    pub struct Isize1;
    impl IterScan for Isize1 {
        type Output = isize;
        #[inline]
        fn scan<'a, I: Iterator<Item = &'a str>>(iter: &mut I) -> Option<Self::Output> {
            isize::scan(iter).map(|x| x.wrapping_sub(1))
        }
    }
}
#[macro_export]
macro_rules !min {($e :expr ) =>{$e } ;($e :expr ,$($es :expr ) ,+) =>{std ::cmp ::min ($e ,min !($($es ) ,+) ) } ;}
#[macro_export]
macro_rules !chmin {($dst :expr ,$($src :expr ) ,+) =>{{let x =std ::cmp ::min ($dst ,min !($($src ) ,+) ) ;$dst =x ;} } ;}
#[macro_export]
macro_rules !max {($e :expr ) =>{$e } ;($e :expr ,$($es :expr ) ,+) =>{std ::cmp ::max ($e ,max !($($es ) ,+) ) } ;}
#[macro_export]
macro_rules !chmax {($dst :expr ,$($src :expr ) ,+) =>{{let x =std ::cmp ::max ($dst ,max !($($src ) ,+) ) ;$dst =x ;} } ;}
fn main() {
    #[allow(unused_imports)]
    use std::io::{Read as _, Write as _};
    let __out = std::io::stdout();
    let mut __in_buf = String::new();
    std::io::stdin().read_to_string(&mut __in_buf).unwrap();
    let mut __scanner = Scanner::new(&__in_buf);
    #[allow(unused_macros)]
    macro_rules !scan {() =>{scan !(usize ) } ;(($($t :tt ) ,*) ) =>{($(scan !($t ) ) ,*) } ;([$t :tt ;$len :expr ] ) =>{(0 ..$len ) .map (|_ |scan !($t ) ) .collect ::<Vec <_ >>() } ;({$t :tt =>$f :expr } ) =>{$f (scan !($t ) ) } ;(chars ) =>{__scanner .scan_chars () } ;($t :ty ) =>{__scanner .scan ::<$t >() } ;}
    let mut __out = std::io::BufWriter::new(__out.lock());
    #[allow(unused_macros)]
    macro_rules !print {($($arg :tt ) *) =>(::std ::write !(__out ,$($arg ) *) .unwrap () ) }
    #[allow(unused_macros)]
    macro_rules !println {($($arg :tt ) *) =>(::std ::writeln !(__out ,$($arg ) *) .unwrap () ) }
    #[allow(unused_macros)]
    macro_rules! echo {
        ($iter :expr ) => {
            echo!($iter, "\n")
        };
        ($iter :expr ,$sep :expr ) => {
            let mut iter = $iter;
            if let Some(item) = iter.next() {
                print!("{}", item);
            }
            for item in iter {
                print!("{}{}", $sep, item);
            }
            println!();
        };
    }
    let n = scan!();
    let s = scan!([i64; n]);
    let mut ans = 0;
    // a>b
    // a-b = c
    // c i + a == n-1
    for c in 1..n - 1 {
        let mut v = vec![false; n];
        v[0] = true;
        v[n - 1] = true;
        let mut sum = 0;
        for x in (c..n).step_by(c) {
            if v[x] || v[n - 1 - x] || x >= n - 1 - x {
                break;
            }
            v[x] = true;
            v[n - 1 - x] = true;
            sum += s[x] + s[n - 1 - x];
            chmax!(ans, sum);
        }
    }
    println!("{}", ans);
}