There are n armchairs, numbered from 1 to n from left to right. Some armchairs are occupied by people (at most one person per armchair), others are not. The number of occupied armchairs is not greater than n/2.

For some reason, you would like to tell people to move from their armchairs to some other ones. If the i-th armchair is occupied by someone and the j-th armchair is not, you can tell the person sitting in the i-th armchair to move to the j-th armchair. The time it takes a person to move from the i-th armchair to the j-th one is |i - j| minutes. You may perform this operation any number of times, but these operations must be done sequentially, i. e. you cannot tell a person to move until the person you asked to move in the last operation has finished moving to their destination armchair.

You want to achieve the following situation: every seat that was initially occupied must be free. What is the minimum time you need to do it?

Input

The first line contains one integer n (2 ≤ n ≤ 5000) — the number of armchairs.

The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1). a_i = 1 means that the i-th armchair is initially occupied, a_i = 0 means that it is initially free. The number of occupied armchairs is at most n/2.

Output

Print one integer — the minimum number of minutes you have to spend to achieve the following situation: every seat that was initially occupied must be free.

Examples

Input


7
1 0 0 1 0 0 1


Output


3


Input


6
1 1 1 0 0 0


Output


9


Input


5
0 0 0 0 0


Output


0

Note

In the first test, you can perform the following sequence:

  1. ask a person to move from armchair 1 to armchair 2, it takes 1 minute; 
  2. ask a person to move from armchair 7 to armchair 6, it takes 1 minute; 
  3. ask a person to move from armchair 4 to armchair 5, it takes 1 minute. 



In the second test, you can perform the following sequence:

  1. ask a person to move from armchair 1 to armchair 4, it takes 3 minutes; 
  2. ask a person to move from armchair 2 to armchair 6, it takes 4 minutes; 
  3. ask a person to move from armchair 4 to armchair 5, it takes 1 minute; 
  4. ask a person to move from armchair 3 to armchair 4, it takes 1 minute. 



In the third test, no seat is occupied so your goal is achieved instantly.
import java.util.*;
import java.lang.*;
// import java.math.*;
import java.io.*;
public class Main
{
    public static int mod=(int)Math.pow(10,9) + 7;
    public static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
    public static PrintWriter ot=new PrintWriter(System.out);
    public static int[] take_arr(int n){
        int a[]=new int[n];
        try {
            String s[]=br.readLine().trim().split(" ");
            for(int i=0;i<n;i++)
            a[i]=Integer.parseInt(s[i]);
        } catch (Exception e) {
            e.printStackTrace();
        }
        return a;
    }
	public static void main (String[] args) throws java.lang.Exception
	{
		try{
            int t=1;//Integer.parseInt(br.readLine().trim());
            int cno=1;
            while(t-->0)
            {
                String qq[]=br.readLine().trim().split(" ");
                int n=Integer.parseInt(qq[0]);
                // int m=Integer.parseInt(qq[1]);
                int a[]=take_arr(n);
                int ans=0;
                for(int i=0;i<n;i++)
                {
                    if(a[i]==1)
                    {

                        int left=-1,right=-1;
                        boolean l=false,r=false;
                        for(int j=i-1;j>-1;j--)
                        {
                            if(a[j]==0)
                            {
                                left=j;
                                l=true;
                                break;
                            }
                        }
                        for(int j=i+1;j<n;j++)
                        {
                            if(a[j]==0)
                            {
                                right=j;
                                r=true;
                                break;
                            }
                        }
                        
                        if(l&&r&&i-left>right-i)
                        {
                            a[right]=2;
                            ans+=right-i;
                            
                        }
                        else if(l&&r&&i-left<right-i)
                        {
                            a[left]=2;
                            ans+=i-left;
                        }
                        else 
                        {
                            if(!r)
                            {
                                a[left]=2;
                                ans+=i-left;
                                // ot.println(i + " " + left);
                            }
                            else
                            {
                                a[right]=2;
                                ans+=right-i;
                            }
                        }
                    }
                }
                ot.println(ans);
            }
            ot.close();
            br.close();
            
        }catch(Exception e){
            e.printStackTrace();
            return;
        }
    }
    
}

    