Virtual contest is a way to take part in past contest, as close as possible to participation on time. It is supported only ACM-ICPC mode for virtual contests.
If you've seen these problems, a virtual contest is not for you - solve these problems in the archive.
If you just want to solve some problem from a contest, a virtual contest is not for you - solve this problem in the archive.
Never use someone else's code, read the tutorials or communicate with other person during a virtual contest.

No tag edit access

E. Product Sum

time limit per test

1 secondmemory limit per test

256 megabytesinput

standard inputoutput

standard outputBlake is the boss of Kris, however, this doesn't spoil their friendship. They often gather at the bar to talk about intriguing problems about maximising some values. This time the problem is really special.

You are given an array *a* of length *n*. The characteristic of this array is the value — the sum of the products of the values *a*_{i} by *i*. One may perform the following operation exactly once: pick some element of the array and move to any position. In particular, it's allowed to move the element to the beginning or to the end of the array. Also, it's allowed to put it back to the initial position. The goal is to get the array with the maximum possible value of characteristic.

Input

The first line of the input contains a single integer *n* (2 ≤ *n* ≤ 200 000) — the size of the array *a*.

The second line contains *n* integers *a*_{i} (1 ≤ *i* ≤ *n*, |*a*_{i}| ≤ 1 000 000) — the elements of the array *a*.

Output

Print a single integer — the maximum possible value of characteristic of *a* that can be obtained by performing no more than one move.

Examples

Input

4

4 3 2 5

Output

39

Input

5

1 1 2 7 1

Output

49

Input

3

1 1 2

Output

9

Note

In the first sample, one may pick the first element and place it before the third (before 5). Thus, the answer will be 3·1 + 2·2 + 4·3 + 5·4 = 39.

In the second sample, one may pick the fifth element of the array and place it before the third. The answer will be 1·1 + 1·2 + 1·3 + 2·4 + 7·5 = 49.

Codeforces (c) Copyright 2010-2017 Mike Mirzayanov

The only programming contests Web 2.0 platform

Server time: Mar/25/2017 03:03:33 (c4).

Desktop version, switch to mobile version.
User lists

Name |
---|