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.
/* 
    Author: Aman Patel
    Date: 16-05-2021
*/


import java.io.*;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class Ed109 {

    static int N = 5000;
    static long[][] dp_arr = new long[N + 5][N + 5];

    //static int _count = 0;
    public static void main(String[] args) {

        FastScanner fs = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);

        for (int i = 0; i <= N; i++) {
            for (int j = 0; j <= N; j++) {
                dp_arr[i][j] = -1;
            }
        }

        int n;
        n = fs.nextInt();

        ArrayList<Integer> occupied = new ArrayList<>(), empty = new ArrayList<>();
        int temp;
        for (int i = 0; i < n; i++) {
            temp = fs.nextInt();
            if (temp == 1)
                occupied.add(i);
            else
                empty.add(i);
        }

        int n1 = occupied.size(), n2 = empty.size();
        long result = cost(occupied, empty, n1, n2, 0, 0);

        out.println(result);
        //out.println(_count);
        out.close();
    }

    static long cost(ArrayList<Integer> occupied, ArrayList<Integer> empty, int n1, int n2, int i, int j) {
        //_count++;
        if (i == n1)
            return 0;
        else if (j == n2)
            return (long) (1e15);

        if (dp_arr[i][j] != -1)
            return dp_arr[i][j];

        long _min = Math.min(Math.abs(occupied.get(i) - empty.get(j)) + cost(occupied, empty, n1, n2, i + 1, j + 1), cost(occupied, empty, n1, n2, i, j + 1));
        dp_arr[i][j] = _min;
        return dp_arr[i][j];
    }

    static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

        public String next() {
            while (!st.hasMoreElements())
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        Long nextLong() {
            return Long.parseLong(next());
        }
    }

    public static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
}
