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.
/*
bts songs to dance to:
I need U
Run
ON
Filter
I'm fine
 */
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import static java.lang.System.out;
import java.util.*;
import java.io.*;
import java.math.*;

public class x1525D
{
    static long INF = Long.MAX_VALUE/2;
    public static void main(String hi[]) throws Exception
    {
        BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(infile.readLine());
        int N = Integer.parseInt(st.nextToken());
        int[] arr = readArr(N, infile, st);
        ArrayList<Integer> zeros = new ArrayList<Integer>();
        ArrayList<Integer> ones = new ArrayList<Integer>();
        for(int i=0; i < N; i++)
        {
            if(arr[i] == 0)
                zeros.add(i);
            else
                ones.add(i);
        }
        if(ones.size() == 0)
        {
            System.out.println(0);
            return;
        }
        N = ones.size();
        int M = zeros.size();
        long[][] dp = new long[N][M];
        for(int i=0; i < N; i++)
            Arrays.fill(dp[i], INF);
        //last move
        for(int b=0; b < M; b++)
        {
            dp[N-1][b] = abs(ones.get(N-1)-zeros.get(b));
        }
        for(int a=N-2; a >= 0; a--)
        {
            for(int b=M-2; b >= 0; b--)
                dp[a+1][b] = min(dp[a+1][b], dp[a+1][b+1]);
            for(int b=0; b < M; b++)
            {
                //if(b > 0)
                 //   dp[a][b] = min(dp[a][b], dp[a][b-1]);
                //take this element
                if(b+1 < M)
                    dp[a][b] = min(dp[a][b], dp[a+1][b+1]+abs(ones.get(a)-zeros.get(b)));
            }
        }
        long res = dp[0][0];
        for(int b=1; b < M; b++)
            res = min(res, dp[0][b]);
        System.out.println(res);
    }
    public static int[] readArr(int N, BufferedReader infile, StringTokenizer st) throws Exception
    {
        int[] arr = new int[N];
        st = new StringTokenizer(infile.readLine());
        for(int i=0; i < N; i++)
            arr[i] = Integer.parseInt(st.nextToken());
        return arr;
    }
    public static void print(ArrayList<Integer> arr)
    {
        //for debugging only
        for(int x: arr)
            out.print(x+" ");
        out.println();
    }
}
/*
dp is likely
overlapping ranges, are they ever optimal?
draw directed graph (two layers)
 */