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.Scanner;
import java.util.ArrayList;

 public class D_Armchairs{
     public static long solve(ArrayList<Integer> zeroes,ArrayList<Integer> ones,int i, int j,long matrix[][],int count){
                if(matrix[i][j]!=0){
                      return matrix[i][j];
                }
                if(count==0){
                    return 0;
                }
                if(i==zeroes.size() || j==ones.size()){
                    long a= Integer.MAX_VALUE;
                    
                    return a;
                }
               
                long ans1= Math.abs(zeroes.get(i)-ones.get(j))+solve(zeroes,ones,i+1,j+1,matrix,count-1);
                long ans2= solve(zeroes,ones,i+1,j,matrix,count);
                long finalans=Math.min(ans1,ans2);
                
                matrix[i][j]=finalans;
                return finalans;
     }
      public static void main(String[] args) {
          Scanner s= new Scanner(System.in);
          int n=s.nextInt();
          ArrayList<Integer> zeroes = new ArrayList<Integer>();
          ArrayList<Integer> ones = new ArrayList<Integer>();
          int array[]= new int[n];
          for(int i=0;i<n;i++){
               array[i]=s.nextInt();
              if(array[i]==0){
                  zeroes.add(i);
              }
              else{
                  ones.add(i);
              }
          }
          long matrix[][]= new long[5002][5002];
          if(ones.size()==0){
            System.out.println(0);
            return;
          }
          int count=ones.size();
          long ans= solve(zeroes,ones,0,0,matrix,count);
          System.out.println(ans);


     }

 }
