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.*;

public class Solution {

    static class FastReader {

        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {

            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

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

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    public static void main(String args[]) {
        FastReader scn = new FastReader();
        int n = scn.nextInt();
        int[] arr = new int[n];

        for (int i = 0; i < n; i++) {
            arr[i] = scn.nextInt();

        }
        ArrayList<Integer> one = new ArrayList<>();
        ArrayList<Integer> two = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            if (arr[i] == 1) {
                one.add(i);
            } else {
                two.add(i);
            }
        }
        int[] arr1 = one.stream().mapToInt(i -> i).toArray();
        int[] arr2 = two.stream().mapToInt(i -> i).toArray();
        int[][] dp = new int[arr1.length + 1][arr2.length + 1];
        for (int[] a : dp) {
            Arrays.fill(a, -1);
        }

        System.out.println(solve(arr1, 0, arr2, 0, dp));
    }

    public static int solve(int[] arr1, int vidx1, int[] arr2, int vidx2, int[][] dp) {
        int n = arr1.length;
        int m = arr2.length;

        if (vidx1 == n) {
            return 0;
        }

        if (m - vidx2 < n - vidx1) {
            return 100000000;
        }
        if (dp[vidx1][vidx2] != -1) {
            return dp[vidx1][vidx2];
        }

        int one = solve(arr1, vidx1 + 1, arr2, vidx2 + 1, dp) + Math.abs(arr1[vidx1] - arr2[vidx2]);
        int two = solve(arr1, vidx1, arr2, vidx2 + 1, dp);

        return dp[vidx1][vidx2] = Math.min(one, two);
    }

}
