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.io.*;

public class Codeforces {
	public static class FastReader {  
        BufferedReader br;
        StringTokenizer st;
 
        public FastReader()
        {
            br = new BufferedReader(
                new InputStreamReader(System.in));
        }
 
        String next()
        {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
 
        int nextInt() { return Integer.parseInt(next()); }
 
        long nextLong() { return Long.parseLong(next()); }
 
        double nextDouble()
        {
            return Double.parseDouble(next());
        }
 
        String nextLine()
        {
            String str = "";
            try {
                str = br.readLine();
            }
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		FastReader sc=new FastReader();
		int n=sc.nextInt();
		int a[]=new int[n];
		ArrayList<Integer> arr0=new ArrayList<>();
		ArrayList<Integer> arr1=new ArrayList<>();
		for(int i=0;i<n;i++)
		{
		    a[i]=sc.nextInt();
		    if(a[i]==0)
		    arr0.add(i);
		    else
		    arr1.add(i);
		}
		n=arr0.size();
		int m=arr1.size();
		int dp[][]=new int[m+1][n+1];
		for(int i=0;i<=n;i++)
		{
		    dp[0][i]=0;
		}
		for(int i=1;i<=m;i++)
		{
		    dp[i][i]=dp[i-1][i-1]+Math.abs(arr0.get(i-1)-arr1.get(i-1));
		    for(int j=i+1;j<=n;j++)
		    {
		        dp[i][j]=Math.min(dp[i-1][j-1]+Math.abs(arr0.get(j-1)-arr1.get(i-1)),dp[i][j-1]);
		    }
		}
		System.out.println(dp[m][n]);
	}
}











