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.io.*;
import java.util.*;
import java.text.*;
import java.math.*;

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());
        StringTokenizer st = new StringTokenizer(br.readLine());
        int ar[] = new int[n];
        for(int i = 0; i < n; i++){
            ar[i] = Integer.parseInt(st.nextToken());
        }
        ArrayList<Integer> ones = new ArrayList<Integer>();
        ArrayList<Integer> zeroes = new ArrayList<Integer>();
        for(int i = 0; i < n; i++){
            if(ar[i] == 1)
                ones.add(i);
            else
                zeroes.add(i);
        }
        int r = ones.size();
        int c = zeroes.size();
        int time[][] = new int[r][c];
        System.out.println(calculateTime(time, r, c, 0, 0, ones, zeroes));
    }
    public static int calculateTime(int time[][], int r, int c, int currR, int currC, ArrayList<Integer> ones, ArrayList<Integer> zeroes){
        // System.out.println(currR + " " + currC);
        if(currR == r)
            return 0;
        if(currC == c)
            return (int)1e9;
        if(time[currR][currC] != 0)
            return time[currR][currC];
        return time[currR][currC] = Math.min((calculateTime(time, r, c, currR + 1, currC + 1, ones, zeroes) + Math.abs(ones.get(currR) - zeroes.get(currC))), calculateTime(time, r, c, currR, currC + 1, ones, zeroes));
    }
}