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.util.*;

import java.io.*;
import  java.lang.Math.*;
public class Solutions {
	 
static int MAX=Integer.MAX_VALUE;
		static int MIN=Integer.MIN_VALUE;
//static ArrayList<ArrayList<Integer>>list=new ArrayList<ArrayList<Integer>>();
	static FastScanner scr=new FastScanner();
	static PrintStream out=new PrintStream(System.out);
//	static ArrayList<Integer>list[];
//	static StringBuilder sb=new StringBuilder();
	public static void main(String []args) {
//		int t=scr.nextInt();
//		while(t-->0) {
			solve();
//		}
//		out.println(sb);
	}
//	static HashMap<Integer,Integer>hm;
	static void solve() {
		int n=scr.nextInt();
		int []a=scr.readArray(n);
		
		int ma=MAX;
		for(int floor=0;floor<n;floor++) {
			int electric=0;
			for(int i=0;i<n;i++) {
				electric+=a[i]*(Math.abs(i-floor)+floor+i);
			}
			ma=Math.min(electric,ma);
		}
		out.println(ma*2);
	}
	static class pair{
		int x;
		int y;
		pair(int x,int y){
			this.x=x;
			this.y=y;
		}
	}
	static boolean isVowel(char c) {
		return c=='a' || c=='e'||c=='i'|| c=='o' || c=='u';
	}
	static int gcd(int a,int b){
		if(b==0) {
			return a;
		}
		
		return gcd(b,a%b);
	}
	static long modPow(long base,long exp,long mod) {
		if(exp==0) {
			return 1;
		}
		if(exp%2==0) {
			long res=(modPow(base,exp/2,mod));
			return (res*res)%mod;
		}
		return (base*modPow(base,exp-1,mod))%mod;
	}
	static long gcd(long a,long b) {
		if(b==0) {
			return a;
		}
		return gcd(b,a%b);
	}
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
	 	String next() {
	 		while (!st.hasMoreTokens())
	 			try {
	 				st=new StringTokenizer(br.readLine());
	 			} catch (IOException e) {
	 				e.printStackTrace();
	 			}
	 			return st.nextToken();
		}
		int[] sort(int a[]) {
			Arrays.sort(a);
			return a;
		}
		int []reverse(int a[]){
			int b[]=new int[a.length];
			int index=0;
			for(int i=a.length-1;i>=0;i--) {
				b[index]=a[i];
			}
			return b;
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long[] readLongArray(int n) {
			long  [] a=new long  [n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}

}
