Codeforces 313C Ilya and Matrix【思维】

C. Ilya and Matrix

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Ilya is a very good-natured lion. He likes maths. Of all mathematical objects, his favourite one is matrices. Now he's faced a complicated matrix problem he needs to solve.

He's got a square 2ⁿ × 2ⁿ-sized matrix and4ⁿ integers. You need to arrange all these numbers in the matrix (put each number in a single individual cell) so that thebeauty of the resulting matrix with numbers is maximum.

The beauty of a 2ⁿ × 2ⁿ-sized matrix is an integer, obtained by the following algorithm:

1. Find the maximum element in the matrix. Let's denote it as m.
2. If n = 0, then the beauty of the matrix equalsm. Otherwise, a matrix can be split into 4 non-intersecting2^n - 1 × 2^n - 1-sized submatrices, then the beauty of the matrix equals the sum of numberm and other four beauties of the described submatrices.

As you can see, the algorithm is recursive.

Help Ilya, solve the problem and print the resulting maximum beauty of the matrix.

Input

The first line contains integer 4ⁿ(1 ≤ 4ⁿ ≤ 2·10⁶). The next line contains4ⁿ integersa_i(1 ≤ a_i ≤ 10⁹) — the numbers you need to arrange in the2ⁿ × 2ⁿ-sized matrix.

Output

On a single line print the maximum value of the beauty of the described matrix.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use thecin, cout streams or the%I64d specifier.

Examples

Input

113

Output

13

Input

41 2 3 4

Output

14

Note

Consider the second sample. You need to arrange the numbers in the matrix as follows:

1 23 4

Then the beauty of the matrix will equal: 4 + 1 + 2 + 3 + 4 = 14.

题目大意：

一个4^n一个矩形，让你将这些数都放在每一个格子中。

定义一个记矩阵的值是这个矩阵中最大价值+一分为四的四个小矩阵的价值。

很显然这是一个递归的过程。

问最大价值。

思路：

我们只要规划出每个数字将会作为矩阵中最大值的次数即可。

很显然，最大值会出现n+1次，然后接下来次大值会出现n次，次次大值也会出现n次，次次次大值也会出现n次，次次次次大值会出现n-1次..........................

依次类推我们可以得到一个规律，统计和即可。

#include<stdio.h>#include<string.h>#include<algorithm>#include<math.h>using namespace std;int a[3000600];int main(){    int n;    while(~scanf("%d",&n))    {        for(int i=1;i<=n;i++)scanf("%d",&a[i]);        sort(a+1,a+n+1);        reverse(a+1,a+1+n);        __int64 output=0;        int now=0;        while(1)        {            int tmp=pow(4,now);            now++;            if(tmp>n)break;            else            {                for(int i=1;i<=tmp;i++)                {                    output+=a[i];                }            }        }        printf("%I64d\n",output);    }}