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: Accepted)

#include <bits/stdc++.h>
using namespace std;
 
using ll = long long;
using ull = unsigned long long;
using ld = long double;
#define REP(NAME, NUM) for (int NAME = 0; NAME < (NUM); ++NAME)
#define BREP(NAME, NUM) for (int NAME = (NUM)-1; NAME >= 0; --NAME)
#define ALL(NAME) (NAME).begin(), (NAME).end()
 
#define cMOD 1000000007ULL
#define cINF ((1ull<<62)-1)
#define cINFINT ((1<<30)-1)

 
template <typename T>
inline vector<T> divisor( T n )
{
	vector<T> res;
	for( T i = 1; i * i <= n; ++i )
	{
		if( n % i == 0 )
		{
			res.push_back( i );
			if( i != n / i ) res.push_back( n / i );
		}
	}
	sort( ALL( res ) );
	return res;
}
 
template <typename T>
inline T gcd(T a, T b)
{
	if (a < b)
		swap(a, b);
	return b ? gcd(b, a % b) : a;
}
 
int main()
{
	ll n; cin >> n;
	vector<ll> s( n, 0 );
	REP(i,n) cin>>s[i];

	vector<vector<ll>> dp(n, vector<ll>());
	REP(i,n) if( i >= 1 )
	{
		dp[i].push_back( 0 );
		for( ll j = 1; j * i < n; ++j )
		{
			auto sum = s[i*j] + s[n-1-i*j] + dp[i][j-1];
			dp[i].push_back( sum );
		}
	}
 
	ll ans = 0;
	REP( a, n )
	{
		if( a <= 2 || a == n - 1 ) continue;
		auto div = divisor<ll>( n - 1 - a );
		for( auto m : div )
		{
			ll b = a - ( n - 1 - a ) / m;
			//cout << a << " " << b << " " << m << " " << endl;
			if( a <= b || b <= 0 ) continue;
 
			// どぼん
			ll g = gcd<ll>( a, b );
			ll a_ = a / g;
			ll b_ = b / g;
			if( a_ == b_ + 1 && b_ < m ) continue;
 
			ll sc = dp[a-b][m];
			//cout << a << " " << b << " " << sc << endl;
			ans = max( ans, sc );
		}
	}
	cout << ans << endl;
 
	return 0;
}