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.*;
import java.lang.*;

public class Solution {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter out = new PrintWriter(System.out);
    StringTokenizer st = new StringTokenizer("");

    String next() {
        if (!st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {

            }
        }
        return st.nextToken();
    }

    int nextInt() {
        return Integer.parseInt(next());
    }

    long nextLong() {
        return Long.parseLong(next());
    }

    public static void main(String[] args) {
        new Solution().solve();
    }

    long res = Long.MAX_VALUE;
    List<Integer> l = new ArrayList<>(), r = new ArrayList<>();
    long dp[][];

    long ok(int i, int j) {
        if (dp[i][j] != -1)
            return dp[i][j];
        if (i >= l.size()) {
            return 0;
        }
        if (j >= r.size())
            return Integer.MAX_VALUE;
        long op1 = ok(i + 1, j + 1) + Math.abs(l.get(i) - r.get(j));
        long op2 = ok(i, j + 1);

        return dp[i][j] = Math.min(op1, op2);
    }

    void solve() {
        int t = 1;
        // t = nextInt();
        for (int tt = 0; tt < t; tt++) {
            int n = nextInt();
            int arr[] = new int[n];
            for (int i = 0; i < n; i++) {
                arr[i] = nextInt();
                if (arr[i] == 0)
                    r.add(i + 1);
                else
                    l.add(i + 1);
            }
            dp = new long[n][n];
            for (int i = 0; i < n; i++)
                Arrays.fill(dp[i], -1);
            out.println(ok(0, 0));
        }
        out.close();
    }
}