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 ArmChairs {
    
 
    public static int solution(int n, int[] arr) {
 
        ArrayList<Integer> one = new ArrayList<Integer>();
        ArrayList<Integer> zero = new ArrayList<Integer>();
 
        for (int i = 0; i < n; i++) {
            if (arr[i] == 1) {
                one.add(i);
            } else {
                zero.add(i);
            }
        }
 
        int[][] dp = new int[one.size() + 1][zero.size() + 1];
 
        for (int i = 1; i <= one.size(); i++) {
            dp[i][i] = dp[i - 1][i - 1] + Math.abs(one.get(i - 1) - zero.get(i - 1));
            for (int j = i + 1; j <= zero.size(); j++) {
                dp[i][j] = Math.min(dp[i][j - 1], dp[i - 1][j - 1] + Math.abs(one.get(i - 1) - zero.get(j - 1)));
            }
        }
 
        return dp[one.size()][zero.size()];
    }
 
 
    public static void main(String[] args) throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter log = new BufferedWriter(new OutputStreamWriter(System.out));
 
        int n = Integer.parseInt(br.readLine());
 
        String[] s = br.readLine().split(" ");
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = Integer.parseInt(s[i]);
        } 
 
        log.write(Integer.toString(solution(n, arr)) + "\n");
        log.flush();
        
    }
 
}