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.io.*;


public class Main{
    
    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];
        String line = br.readLine();
        String[] parts = line.trim().split("\\s+");
        for(int i=0; i<n; i++)
        a[i] = Integer.parseInt(parts[i]);
        
        boolean[] vis = new boolean[n];
        
        int ans = 0;
        for(int i=0; i<n; i++)
        {
            if(a[i] == 1 && !vis[i])
            {
                vis[i] = true;
                
                int left_d = Integer.MAX_VALUE, right_d = Integer.MAX_VALUE;
                if(i == 0){
                    for(int right = i+1; right<n; right++)
                    {
                        if(a[right] == 0 && !vis[right]){
                            vis[right] = true;
                            right_d = right;
                            break;
                        }
                        
                    }
                    a[i] = 0;
                    a[right_d] = 1;
                    ans += (right_d-i);
                }
                else if(i == n-1)
                {
                    for(int left = i-1; left>=0; left--)
                    {
                        if(a[left] == 0 && !vis[left]){
                            vis[left] = true;
                            left_d = left;
                            break;
                        } 
                    }
                    a[i] = 0;
                    a[left_d] = 1;
                    ans += Math.abs(left_d-i);
                }
                
                else{
                    for(int right = i+1; right<n; right++)
                    {
                        if(a[right] == 0 && !vis[right]){
                            right_d = right;
                            break;
                        }
                        
                    }
                    
                    for(int left = i-1; left>=0; left--)
                    {
                        if(a[left] == 0 && !vis[left]){
                            left_d = left;
                            break;
                        } 
                    }
                    
                    if(left_d < right_d){
                        ans += Math.abs(left_d-i);
                        vis[left_d] = true;
                        a[i] = 0;
                        a[left_d] = 1;
                    }
                    else{
                        ans += Math.abs(right_d-i);
                        vis[right_d] = true;
                        a[i] = 0;
                        a[right_d] = 1;
                    }
                    
                    
                }
                
            }
            
        }
        System.out.println(ans);

    }
}