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.*;
import java.io.*;

public class _1525_D {
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);
        int n = Integer.parseInt(in.readLine());
        int[] a = new int[n];
        StringTokenizer line = new StringTokenizer(in.readLine());
        int c = 0;
        for(int i = 0; i < n; i++) {
            a[i] = Integer.parseInt(line.nextToken());
            if(a[i] == 1) c++;
        }
        int[] pos = new int[c];
        int count = 0;
        for(int i = 0; i < n; i++) {
            if(a[i] == 1) {
                pos[count] = i;
                count++;
            }
        }
        int[][] dp = new int[c + 1][n];
        for(int i = 1; i <= c; i++) {
            Arrays.fill(dp[i], Integer.MAX_VALUE);
        }
        for(int i = 1; i <= c; i++) {
            for(int j = 0; j < n; j++) {
                if(j > 0) {
                    if(a[j] == 0) {
                        if(dp[i - 1][j - 1] != Integer.MAX_VALUE) {
                            dp[i][j] = dp[i - 1][j - 1] + Math.abs(j - pos[i - 1]);
                        }
                    }
                    dp[i][j] = Math.min(dp[i][j], dp[i][j - 1]);
                }else {
                    if(a[j] == 0) {
                        dp[i][j] = Math.abs(j - pos[i - 1]);
                    }
                }
            }
        }
        out.println(dp[c][n - 1]);
        in.close();
        out.close();
    }
}