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 {
    static Reader in = new Reader(System.in);
    static StringBuilder stringBuilder = new StringBuilder();
    public static void main(String[] args) {
        int t = 1;
        while (t-- > 0) {
            int n = in.nextInt();
            int[] a = new int[n];
            int count = 0;
            for (int i = 0; i < n; i++) {
                a[i] = in.nextInt();
                if (a[i] == 1) {
                    count ++;
                }
            }
            int[] occ = new int[count];
            int[] empty = new int[n - count];
            int c = 0, c1 = 0;
            for (int i = 0; i < n; i++) {
                if (a[i] == 1) {
                    occ[c++] = i;
                } else {
                    empty[c1++] = i;
                }
            }
            n = occ.length;
            int m = empty.length;
            int[][] dp = new int[n + 1][m + 1];
            int[][] min = new int[n + 1][m + 1];
            int inf = Integer.MAX_VALUE / 55;
            for (int i = 0; i <= n; i++) {
                Arrays.fill(dp[i], inf);
                Arrays.fill(min[i], inf);
            }
            for (int i = 0; i <= m; i++) {
                dp[0][i] = 0;
                min[0][i] = 0;
            }
            for (int i = 1; i <= n; i++) {
                for (int j = 1; j <= m; j++) {
                    dp[i][j] = Math.abs(occ[i - 1] - empty[j - 1]) + min[i - 1][j - 1];
                    min[i][j] = Math.min(min[i][j - 1], dp[i][j]);
                }
            }
            stringBuilder.append(min[n][m]).append("\n");
        }
        System.out.println(stringBuilder);
    }
}
class Reader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;

    public Reader(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());
    }

}