#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=2e5+7;
ll s[maxn];
ll f[maxn];
int main()
{
	ll n;cin>>n;
	for(ll i=0;i<n;i++)
	{
		scanf("%lld",&s[i]);
	}
	//ll c=1;
	ll mx=0;
	for(ll c=1;c<n-1;c++)
	{
		ll sz=(n-2)/c;
		unordered_map<ll,ll> mp;
		ll ans=0;
		for(ll k=1;k<=sz;k++)
		{
			if(n-1-k*c-c<=0) break;
			//n-1>=k*c+a,a-b=c>0,a>c
			//n-1-k*c-c should be >0; 
			if(mp.count(n-1-k*c)) break;
			mp[n-1-k*c]++;
			if(mp.count(k*c)) break;
			mp[k*c]++;
			ans+=s[k*c]+s[n-1-k*c];
			mx=max(ans,mx);
		 } 
	}
	cout<<mx<<endl;
}
### Prompt

Develop a solution in CPP to the problem described below:

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.

Examples

Input

5
0 2 5 1 0


Output

3


Input

6
0 10 -7 -4 -13 0


Output

0


Input

11
0 -4 0 -99 31 14 -15 -39 43 18 0


Output

59

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=2e5+7;
ll s[maxn];
ll f[maxn];
int main()
{
	ll n;cin>>n;
	for(ll i=0;i<n;i++)
	{
		scanf("%lld",&s[i]);
	}
	//ll c=1;
	ll mx=0;
	for(ll c=1;c<n-1;c++)
	{
		ll sz=(n-2)/c;
		unordered_map<ll,ll> mp;
		ll ans=0;
		for(ll k=1;k<=sz;k++)
		{
			if(n-1-k*c-c<=0) break;
			//n-1>=k*c+a,a-b=c>0,a>c
			//n-1-k*c-c should be >0; 
			if(mp.count(n-1-k*c)) break;
			mp[n-1-k*c]++;
			if(mp.count(k*c)) break;
			mp[k*c]++;
			ans+=s[k*c]+s[n-1-k*c];
			mx=max(ans,mx);
		 } 
	}
	cout<<mx<<endl;
}
```