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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;

public class _109D {
    static BufferedReader br;

    public static void main(String[] args) throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        int n = readInt();
        int arr[] = readIntarray();
        ArrayList<Integer> a = new ArrayList<>();
        ArrayList<Integer> b = new ArrayList<>();
        for(int i = 0; i < n; i++){
            if(arr[i] == 1){
                a.add(i);
            }else{
                b.add(i);
            }
        }
        if(a.size() == 0){
            System.out.println("0");
            return;
        }
        int [][] dp = new int[a.size()][b.size()];
        for(int i = 0; i < a.size(); i++){
            for(int j = i; j < b.size(); j++) {
                if (j == 0) {
                    dp[i][j] = Math.abs(a.get(i) - b.get(j));
                } else if (i == 0) {
                    dp[i][j] = Math.min(dp[i][j - 1], Math.abs(a.get(i) - b.get(j)));
                } else if (i == j) {
                    dp[i][j] = dp[i - 1][j - 1] + Math.abs(a.get(i) - b.get(j));
                } else {
                    dp[i][j] = Math.min(dp[i][j - 1], dp[i - 1][j - 1] + Math.abs(a.get(i) - b.get(j)));
                }
            }
        }
        System.out.println(dp[a.size() - 1][b.size() - 1]);
    }


    static int readInt() throws IOException {
        return Integer.parseInt(br.readLine());
    }

    static long readLong() throws IOException {
        return Long.parseLong(br.readLine());
    }

    static int[] readIntarray() throws IOException {
        String[] _a = br.readLine().split(" ");
        int[] _res = new int[_a.length];
        for (int i = 0; i < _a.length; i++) {
            _res[i] = Integer.parseInt(_a[i]);
        }
        return _res;
    }

    static long[] readLongarray() throws IOException {
        String[] _a = br.readLine().split(" ");
        long[] _res = new long[_a.length];
        for (int i = 0; i < _a.length; i++) {
            _res[i] = Long.parseLong(_a[i]);
        }
        return _res;
    }
}
