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 Main {
    static class FastInput {
        BufferedReader br;
        StringTokenizer st;

        public FastInput() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

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

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    public static long gcd(long a, long b) {
        if (b > a)
            return gcd(b, a);
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }

    public static void plist(ArrayList<Integer> arr) {
        for (int i = 0; i < arr.size(); i++) {
            System.out.print(arr.get(i) + " ");
        }
        System.out.println();
    }
    static long[][] dp = new long[5005][5005];
    static ArrayList<Integer> filled = new ArrayList<>();
    static ArrayList<Integer> vacant = new ArrayList<>();
    public static long calc(int i, int j){
        if(i >= filled.size())
            return 0;
        if(j >= vacant.size())
            return Integer.MAX_VALUE;
        if(dp[i][j] != -1)
            return dp[i][j];
        dp[i][j] = Math.min(Math.abs(filled.get(i) - vacant.get(j)) + calc(i+1, j+1), calc(i, j+1));
        return dp[i][j];
    }
    
    public static void run_case(FastInput s) {
        int n = s.nextInt();
        int[] arr = new int[n];
        for(int i = 0; i < n; i++){
            arr[i] = s.nextInt();
            if(arr[i] == 1){
                filled.add(i);
            } else {
                vacant.add(i);
            }
        }
        for(int i = 0; i < filled.size(); i++){
            for(int j = 0; j < vacant.size(); j++)
                dp[i][j] = -1;
        }
        System.out.println(calc(0, 0));
    }

    public static void main(String[] args) {
        FastInput s = new FastInput();
        run_case(s);
    }
}