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.ArrayList;
import java.util.List;
import java.util.StringTokenizer;

public class D {
    public static void main(String[] args) {
        new D().solve(System.in, System.out);
    }

    public void solve(InputStream in, OutputStream out) {
        InputReader inputReader = new InputReader(in);
        PrintWriter writer = new PrintWriter(new BufferedOutputStream(out));

        int n = inputReader.nextInt();
        boolean[] occupied = new boolean[n];
        for (int i = 0; i < n; i++) {
            occupied[i] = inputReader.nextInt() == 1;
        }
        writer.println(solve(n, occupied));

        writer.close();
    }

    public int solve(int n, boolean[] occupiedArray) {
        List<Integer> occ = new ArrayList<>();
        List<Integer> free = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            if (occupiedArray[i]) {
                occ.add(i);
            } else {
                free.add(i);
            }
        }
        int occSize = occ.size();
        if (occSize == 0) {
            return 0;
        }
        int freeSize = free.size();
        long[][] d = new long[occSize][freeSize];
        for (int j = 0; j < freeSize; j++) {
            d[0][j] = Integer.MAX_VALUE;
            for (int k = 0; k <= j; k++) {
                d[0][j] = Math.min(d[0][j], Math.abs(occ.get(0) - free.get(k)));
            }
        }

        for (int i = 1; i < occSize; i++) {
            for (int j = 0; j < freeSize; j++) {
                fillIJ(occ, free, d, i, j);
            }
        }
        return (int) d[occSize - 1][freeSize - 1];
    }

    private void fillIJ(List<Integer> occ, List<Integer> free, long[][] d, int i, int j) {
        int left = i;
        int right = j;
        while (left < right) {
            int mid = (left + right) / 2;
            long countForMid = countFor(occ, free, d, i, mid);
            long countForMid1 = countFor(occ, free, d, i, mid + 1);
            if (countForMid1 > countForMid) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        d[i][j] = countFor(occ, free, d, i, left);
    }

    private long countFor(List<Integer> occ, List<Integer> free, long[][] d, int i, int k) {
        return Math.abs(occ.get(i) - free.get(k)) + d[i - 1][k - 1];
    }

    private static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

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

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

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