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.ArrayList;
import java.util.Arrays;
import java.util.Scanner;

/**
 *
 * @author user
 */
public class Armchair {
    public static void main(String args[]){
        Scanner in=new Scanner(System.in);
        int n=in.nextInt();
        int arr[]=new int[n];
        ArrayList<Integer> list1=new ArrayList<Integer>();
         ArrayList<Integer> list2=new ArrayList<Integer>();
        for(int i=0;i<n;i++)
        {
            int a=in.nextInt();
            if(a==0)
                list2.add(i);
            else
                list1.add(i);
        }
        long dp[][]=new long[list1.size()+1][list2.size()+1];
        
        solve(list1,list2,dp);
        System.out.println(dp[list1.size()][list2.size()]);
    }

public static void solve( ArrayList<Integer> list1,ArrayList<Integer> list2,long dp[][]){
       for(int i=1;i<=list1.size();i++)
		dp[i][0]=Integer.MAX_VALUE;

	for(int i=1;i<=list1.size();i++){
		for(int j=1;j<=list2.size();j++){
			dp[i][j]=Math.min(Math.abs(list1.get(i-1)-list2.get(j-1))+dp[i-1][j-1],dp[i][j-1]);
		}
	}
}
}