The Fair Nut lives in n story house. a_i people live on the i-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 x-th floor, but x hasn't been chosen yet. When a person needs to get from floor a to floor b, elevator follows the simple algorithm: 

  * Moves from the x-th floor (initially it stays on the x-th floor) to the a-th and takes the passenger. 
  * Moves from the a-th floor to the b-th floor and lets out the passenger (if a equals b, elevator just opens and closes the doors, but still comes to the floor from the x-th floor). 
  * Moves from the b-th floor back to the x-th. 

The elevator never transposes more than one person and always goes back to the floor x before transposing a next passenger. The elevator spends one unit of electricity to move between neighboring floors. So moving from the a-th floor to the b-th floor requires |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 x-th floor. Don't forget than elevator initially stays on the x-th floor. 

Input

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

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 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

16

Input

2
1 1


Output

4

Note

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

In the second example, the answer can be achieved by choosing the first floor as the x-th floor.
import java.io.*;
import java.util.*;
import java.math.*;
public 
class Main{
    //static variable
    static final int mod = (int) 1e9 + 7;
    static final double eps = 1e-6;
    static final double pi = Math.PI;
    static final long inf = Long.MAX_VALUE / 2;

    // .......static class
  static class Pair{
      int key,value;
      Pair(int key,int value){
        this.key=key;
        this.value=value;
      }
      int Key(){
        return key;
      }
      int Value(){
        return value;
      }
  }
  static class mycomparator implements Comparator<Pair>{
     @Override
     public int compare(Pair o1,Pair o2){
      Integer key1=o1.Key(),key2=o2.Key();
          return key1.compareTo(key2);
     }
  }
    
//.............staic class end.
  
    BufferedReader br;
    PrintWriter out;
    public static void main(String[] args) {
    new Main().main1();
 
  }
  void main1()
  {
    try{
       
        br=new BufferedReader(new InputStreamReader(System.in));
        out=new PrintWriter(System.out);
        int t=1;
       // t=ii();
      while(t-->0){

        //........solution startn

        int n = ii();
        int[] a = iint();

        int min = Integer.MAX_VALUE;

        int floor = 0;
        for(int i=1;i<=n;i++){
              int temp = 0;
             for(int j = 1;j<=n;j++){
                int cost = 0;
              
                cost = 2*Math.abs(i-j) + 2*(j-1) + 2*(i-1);

               temp = temp + a[j-1]*cost;
             }
             if(min>temp){
              min = temp;
              floor = i;
             }
        }
        System.out.println(min);
        // System.out.println(floor);


      

  
        









        //..........solution end.

     }
     out.flush();
     out.close(); 
    }
    catch(Exception e){
        e.printStackTrace();}    
  }


  // ...............required method.
  int gcd(int a, int b) 
{ 
    if (a == 0) 
        return b; 
    return gcd(b%a, a); 
} 




  //................end.
  
  
   //..............input method start. 
    int getmax(int a[]){
      int n=a.length;
      int max=a[0];
      for (int i=1;i<n ;i++ ) {
        max=Math.max(a[i],max);
      }
      return max;
    }
     int getmin(int a[]){
      int n=a.length;
      int min=a[0];
      for (int i=1;i<n ;i++ ) {
        min=Math.min(a[i],min);
      }
      return min;
    }

    int[] iint() throws IOException{
      String line[]=br.readLine().split(" ");
      int[] a=new int[line.length];
      for (int i=0;i<line.length ;i++ ) {
        a[i]=Integer.parseInt(line[i]);
      }
      return a;
    }
    long[] ilong() throws IOException{
      String line[]=br.readLine().split(" ");
      long[] a=new long[line.length];
      for (int i=0;i<line.length ;i++ ) {
        a[i]=Long.parseLong(line[i]);
      }
      return a;
    }
    double[] idouble() throws IOException{
      String line[]=br.readLine().split(" ");
      double[] a=new double[line.length];
      for (int i=0;i<line.length ;i++ ) {
        a[i]=Long.parseLong(line[i]);
      }
      return a;
    }
    long li() throws IOException{
      
      return Long.parseLong(br.readLine());
    }
    int ii() throws IOException{
      
      return Integer.parseInt(br.readLine());
    }
    double di() throws IOException{
      return Double.parseDouble(br.readLine());
    }
    char ci() throws IOException{
      return (char)br.read();
    }
    String si() throws IOException{
       return br.readLine();
    }
    String[] isa(int n) throws IOException{
        String at =si();
        return at.split(" ");
    }

    double[][] idm(int n, int m) throws IOException{
        double a[][] = new double[n][m];
        for (int i = 0; i < n; i++) {
          double[] temp=idouble();
          for (int j = 0; j < m; j++) a[i][j] = temp[j];
        }
        return a;
    }

    int[][] iim(int n, int m) throws IOException{
        int a[][] = new int[n][m];
        for (int i = 0; i < n; i++) {
          int[] temp=iint();
          for (int j = 0; j < m; j++) a[i][j] =temp[j];
        }
        return a;
    }
    long[][] ilm(int n, int m) throws IOException{
       long a[][] = new long[n][m];
        for (int i = 0; i < n; i++) {
          long[] temp=ilong();
          for (int j = 0; j < m; j++) a[i][j] =temp[j];
        }
        return a;
    }
    //..............input methods end;
    
} 