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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.TreeSet;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        DArmchairs solver = new DArmchairs();
        solver.solve(1, in, out);
        out.close();
    }

    static class DArmchairs {
        int n;
        int[] arr;
        int[][] memo;
        TreeSet<Integer>[] sets;

        public void readInput(Scanner sc) {
            n = sc.nextInt();
            arr = new int[n];
            for (int i = 0; i < n; i++)
                arr[i] = sc.nextInt();
        }

        public void solve(int testNumber, Scanner sc, PrintWriter pw) {
            int tc = 1;
            while (tc-- > 0) {
                readInput(sc);
                sets = new TreeSet[2];
                for (int i = 0; i < 2; i++)
                    sets[i] = new TreeSet<>();
                for (int i = 0; i < n; i++)
                    sets[arr[i]].add(i);
                memo = new int[n][n];
                for (int[] x : memo)
                    Arrays.fill(x, -1);
                if (sets[1].isEmpty()) {
                    pw.println(0);
                } else {
                    pw.println(dp(sets[0].first(), sets[1].first()));
                }
            }
        }

        private int dp(Integer i, Integer j) {
            if (j == null)
                return 0;
            if (j == null || i == null)
                return (int) 1e9;
            if (memo[i][j] != -1)
                return memo[i][j];
            int leave = dp(sets[0].higher(i), j);
            return memo[i][j] = Math.min(Math.abs(j - i) + dp(sets[0].higher(i), sets[1].higher(j)), leave);
        }

    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public String next() {
            try {
                while (st == null || !st.hasMoreTokens())
                    st = new StringTokenizer(br.readLine());
                return st.nextToken();
            } catch (Exception e) {
                throw new RuntimeException(e);
            }
        }

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

    }
}

