Codeforces Round #526 (Div. 2) A

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本文探讨了一种电梯能耗最小化的算法，通过选择最优楼层作为电梯待机位置，以减少日常运营中的电力消耗。通过对每层楼居住人数的考虑，算法遍历所有可能的待机楼层，计算总能耗并选取最小值。

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A. The Fair Nut and Elevator

The Fair Nut lives in nn story house. aiai people live on the ii-th floor of the house. Every person uses elevator twice a day: to get from the floor where he/she lives to the ground (first) floor and to get from the first floor to the floor where he/she lives, when he/she comes back home in the evening.

It was decided that elevator, when it is not used, will stay on the xx-th floor, but xx hasn't been chosen yet. When a person needs to get from floor aa to floor bb, elevator follows the simple algorithm:

Moves from the xx-th floor (initially it stays on the xx-th floor) to the aa-th and takes the passenger.
Moves from the aa-th floor to the bb-th floor and lets out the passenger (if aa equals bb, elevator just opens and closes the doors, but stillcomes to the floor from the xx-th floor).
Moves from the bb-th floor back to the xx-th.

The elevator never transposes more than one person and always goes back to the floor xx before transposing a next passenger. The elevator spends one unit of electricity to move between neighboring floors. So moving from the aa-th floor to the bb-th floor requires |a−b||a−b|units of electricity.

Your task is to help Nut to find the minimum number of electricity units, that it would be enough for one day, by choosing an optimal the xx-th floor. Don't forget than elevator initially stays on the xx-th floor.

Input

The first line contains one integer nn (1≤n≤1001≤n≤100) — the number of floors.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤1000≤ai≤100) — the number of people on each floor.

Output

In a single line, print the answer to the problem — the minimum number of electricity units.

Examples

input

3
0 2 1

output

input

2
1 1

output

Note

In the first example, the answer can be achieved by choosing the second floor as the xx-th floor. Each person from the second floor (there are two of them) would spend 44 units of electricity per day (22 to get down and 22 to get up), and one person from the third would spend 88units of electricity per day (44 to get down and 44 to get up). 4⋅2+8⋅1=164⋅2+8⋅1=16.

不看Note不会做题系列，看了半天才发现a数组里存的是每层多少人，由于数据量小，直接暴力就好了

#include<iostream>
#include<memory.h> 
#include<algorithm>
#include<cmath>
using namespace std;
const int maxn = 1e5+5;
int num[maxn] = {0}; //每层有多少人 
int n, ans = 1000000000;
int main()
{
	cin >> n;
	for(int i = 1;i <= n;i ++)
		cin >> num[i];
	for(int i = 1;i <= n;i ++)
	{
		int cnt = 0;
		for(int j = 1;j <= n;j ++)
			cnt += 2 * num[j] * (abs(i - j) + abs(j - 1) + abs(1 - i));
		ans = min(cnt, ans);
	}
	cout << ans << endl;
	return 0;
}