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.Arrays;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();

        int[] a = new int[n];

        for(int i=0; i<n; i++) {
            a[i] = in.nextInt();
        }

        System.out.println(getMinimumTime(a, n));
    }

    public static int getMinimumTime(int[] a, int n) {
        int[][] minTime = new int[n][n];
        int[] posOfPerson = new int[n];

        int totalOccupied = 0;
        for(int i=0; i<n; i++) {
            if(a[i] == 1) {
                posOfPerson[totalOccupied] = i;
            }
            totalOccupied += a[i];
        }

        for(int i=0; i<n; i++) {
            Arrays.fill(minTime[i], 1, n, Integer.MAX_VALUE);
        }

        if(a[0] == 0) {
            minTime[0][1] = Math.abs(0-posOfPerson[0]);
        }

        int unoccupiedPlaces = 1-a[0];
        for(int i=1; i<n; i++) {
            if(a[i] == 0) {
                unoccupiedPlaces++;
            }

            for(int j=1; j<=Math.min(totalOccupied, unoccupiedPlaces); j++) {
                if(a[i] == 1) {
                    minTime[i][j] = minTime[i-1][j];
                }
                else if(j == unoccupiedPlaces) {
                    minTime[i][j] = minTime[i-1][j-1] + Math.abs(i-posOfPerson[j-1]);
                }
                else {
                    minTime[i][j] = Math.min(minTime[i-1][j], minTime[i-1][j-1] + Math.abs(i-posOfPerson[j-1]));
                }
            }
        }

        return minTime[n-1][totalOccupied];
    }
}
