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.
// package FinalGrind;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class Problem9 {
    public static void main(String[] args)throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int n=Integer.parseInt(br.readLine());
        int a[]=new int[n+1];
        StringTokenizer st=new StringTokenizer(br.readLine());
        ArrayList<Integer> ones=new ArrayList<>();
        ArrayList<Integer> zeros=new ArrayList<>();
        for(int i=1;i<=n;i++){
            a[i]=Integer.parseInt(st.nextToken());
            if(a[i]==1){
                ones.add(i);
            }
            else{
                zeros.add(i);
            }
        }
        if(ones.size()==0){
            System.out.println(0);
            return;
        }
        int dp[]=new int[zeros.size()];
        for(int i=0;i<dp.length;i++){
            int time=Math.abs(zeros.get(i)-ones.get(0));
            if(i==0){
                dp[i]=time;
            }
            else{
                dp[i]=Math.min(dp[i-1],time);
            }
        }
        for(int i=1;i<ones.size();i++){
            int temp[]=dp.clone();
            for(int j=i;j<zeros.size();j++){
                int time=Math.abs(ones.get(i)-zeros.get(j));
                if(j==i){
                    dp[j]=temp[j-1]+time;
                }
                else{
                    dp[j]=Math.min(dp[j-1],temp[j-1]+time);
                }
            }
        }
        System.out.println(dp[dp.length-1]);
    }
}
