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.*;

public class Armchairs {

	public static int findMinTime(List<Integer> zeros, List<Integer> ones) {
		if (ones.size() == 0)
			return 0;
		int oneSize = ones.size();
		int zeroSize = zeros.size();
		int [][] time = new int [oneSize + 1][zeroSize + 1];
		for (int i=1; i<=oneSize; i++) {
			time[i][i] = time[i - 1][i - 1] + Math.abs(ones.get(i - 1) - zeros.get(i - 1));
			for (int j=i+1; j<=zeroSize; j++) {
				time[i][j] = Math.min(time[i][j - 1], time[i - 1][j - 1] + 
					Math.abs(ones.get(i - 1) - zeros.get(j - 1)));
			}
		}
		return time[oneSize][zeroSize];
	}

	public static void main (String [] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		List<Integer> zeros = new ArrayList<>();
		List<Integer> ones = new ArrayList<>();
		for (int i=0; i<n; i++) {
			int number= sc.nextInt();
			if (number == 1)
				ones.add(i);
			else
				zeros.add(i);
		}
		System.out.println(findMinTime(zeros, ones));
	}
}