B. Journey Planning

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本文介绍了Tanya在Berland城市旅行的计划。她选择一个起点，并按递增顺序访问城市，每个相邻城市间的美丽值差必须相等。任务是帮助Tanya规划旅程，使其总美丽值最大。输入包含城市数量和每个城市的美丽值，输出是最大可能的美丽值。给出了几个示例以说明如何确定最佳旅程。

链接：https://codeforces.ml/contest/1321/problem/B

Tanya wants to go on a journey across the cities of Berland. There are nn cities situated along the main railroad line of Berland, and these cities are numbered from 11 to nn.

Tanya plans her journey as follows. First of all, she will choose some city c1c1 to start her journey. She will visit it, and after that go to some other city c2>c1c2>c1, then to some other city c3>c2c3>c2, and so on, until she chooses to end her journey in some city ck>ck−1ck>ck−1. So, the sequence of visited cities [c1,c2,…,ck][c1,c2,…,ck] should be strictly increasing.

There are some additional constraints on the sequence of cities Tanya visits. Each city ii has a beauty value bibi associated with it. If there is only one city in Tanya's journey, these beauty values imply no additional constraints. But if there are multiple cities in the sequence, then for any pair of adjacent cities cici and ci+1ci+1, the condition ci+1−ci=bci+1−bcici+1−ci=bci+1−bci must hold.

For example, if n=8n=8 and b=[3,4,4,6,6,7,8,9]b=[3,4,4,6,6,7,8,9], there are several three possible ways to plan a journey:

c=[1,2,4]c=[1,2,4];
c=[3,5,6,8]c=[3,5,6,8];
c=[7]c=[7] (a journey consisting of one city is also valid).

There are some additional ways to plan a journey that are not listed above.

Tanya wants her journey to be as beautiful as possible. The beauty value of the whole journey is the sum of beauty values over all visited cities. Can you help her to choose the optimal plan, that is, to maximize the beauty value of the journey?

Input

The first line contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the number of cities in Berland.

The second line contains nn integers b1b1, b2b2, ..., bnbn (1≤bi≤4⋅1051≤bi≤4⋅105), where bibi is the beauty value of the ii-th city.

Output

Print one integer — the maximum beauty of a journey Tanya can choose.

Examples

input

Copy

6
10 7 1 9 10 15

output

Copy

input

Copy

1
400000

output

Copy

input

Copy

7
8 9 26 11 12 29 14

output

Copy

Note

The optimal journey plan in the first example is c=[2,4,5]c=[2,4,5].

The optimal journey plan in the second example is c=[1]c=[1].

The optimal journey plan in the third example is c=[3,6]c=[3,6].

又双叒掉到负数的坑里了

#include <bits/stdc++.h>
using namespace std;
long long n,t,k,s;
long long a[2000001],b[2000001];
map<long long,long long>m;
int main()
{
    cin>>n;
    s=0;
    for(int i=1;i<=n;i++)
    {
    	cin>>b[i];
    	m[b[i]-i]+=b[i];
    }
    for(int i=-400000;i<=400000;i++)
    {
    	s=max(m[i],s);
    }
    cout<<s;
    
}