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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;


public class TaskD {

    public static void main(String[] arg) {
        final FastScanner in = new FastScanner(System.in);
        final PrintWriter out = new PrintWriter(System.out);
        final int n = in.nextInt();
        final int[] a = new int[n];
        int zeroCount = 0;
        for(int i=0;i<n;i++){
            a[i]=in.nextInt();
            if(a[i]==0){
                zeroCount++;
            }
        }
        int [] aj = new int[zeroCount];
        int [] ai = new int[n-zeroCount];
        {
            int ii=0;
            int jj=0;
            for(int i=0;i<n;i++){
                if(a[i]==0){
                    aj[jj]=i;
                    jj++;
                }else{
                    ai[ii]=i;
                    ii++;
                }
            }
        }
        int solution = solution(ai,aj);

        out.println(solution);
        out.flush();
        out.close();
        in.close();
    }

    private static int solution(final int[] ai,final int[] aj) {
        int [][] dp = new int[ai.length+1][aj.length+1];
        dp[0][0]=0;
        for(int i=0;i<aj.length;i++){
            dp[0][i]=0;
        }
        final int INF = 1_000_000_000;
        for(int i=1;i<=ai.length;i++){
            for(int j=0;j<i;j++){
                dp[i][j]=INF;
            }
            for(int j=i;j<=aj.length;j++){
                dp[i][j]=Math.min(dp[i-1][j-1]+Math.abs(aj[j-1]-ai[i-1]), dp[i][j-1]);
            }
        }

        return dp[ai.length][aj.length];
    }


    private static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner(InputStream stream) {
            try {
                br = new BufferedReader(new InputStreamReader(stream));
            } catch (Exception e) {
                e.printStackTrace();
            }
        }

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

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

        int[] readIntArr(int n) {
            int[] result = new int[n];
            for (int i = 0; i < n; i++) {
                result[i] = Integer.parseInt(next());
            }
            return result;
        }

        long[] readLongArr(int n) {
            long[] result = new long[n];
            for (int i = 0; i < n; i++) {
                result[i] = Long.parseLong(next());
            }
            return result;
        }

        void close() {
            try {
                br.close();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }

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

    }

}