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 void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt();
		int[] seats = new int[n];
		ArrayList<Integer> occupied = new ArrayList<>();
		for (int i=0;i<n;i++) {
			int temp = scan.nextInt();
			seats[i] = temp;
			if (temp == 1) occupied.add(i);
		}
		int occupiedAtFirst = occupied.size();
		int[][] dp = new int[n+1][occupiedAtFirst+1]; //look at first i seats and choose j seats to be seated on
		for (int i=0;i<=n;i++) {
			for (int j=0;j<=occupiedAtFirst;j++) {
				dp[i][j] = Integer.MAX_VALUE;
			}
		}
		dp[0][0] = 0;
		for (int i=0;i<n;i++) {
			for (int j=0;j<=occupiedAtFirst;j++) {
				int prev = seats[i];
				if (dp[i][j] == Integer.MAX_VALUE) continue; //important! because this means that we have not solved for that value yet...
				if (prev == 1) { //if my current seat was occupied previously
					dp[i+1][j] = Math.min(dp[i+1][j], dp[i][j]); //can only move on without choosing
				}
				else {
					dp[i+1][j] = Math.min(dp[i+1][j], dp[i][j]);
					if (j<occupiedAtFirst) dp[i+1][j+1] = Math.min(dp[i+1][j+1], dp[i][j] + Math.abs(occupied.get(j) - i));
				}
			}
		}
		System.out.println(dp[n][occupiedAtFirst]);
	}
}
