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 Ishu
{
    static Scanner scan = new Scanner(System.in);
    static BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out)); 
    static void tc() throws Exception
    {
    int n = scan.nextInt();
    int[] a = new int[n];
    int one = 0;
    int i,j;
    
    for(i=0;i<n;++i)
        {
        a[i] = scan.nextInt();
        one += a[i];
        }
        
    int[] pos = new int[one];
    one = 0;
    for(i=0;i<n;++i)  
        if(a[i] == 1)
            pos[one++] = i;
          
    long[][] dp = new long[one][n];
    
    if(one == 0)
        {
        output.write("0\n");
        output.flush();
        return;
        }
        
    if(a[0] == 0)
        dp[0][0] = pos[0];
    else 
        dp[0][0] = Integer.MAX_VALUE;
    for(j=1;j<n;++j)    
        {
        if(a[j] == 1)
            {
            dp[0][j] = dp[0][j-1];
            continue;
            }
            
        dp[0][j] = Math.abs(pos[0] - j);
        dp[0][j] = Math.min(dp[0][j], dp[0][j-1]);    
        }
     
     
    for(i=1;i<one;++i)  
        {
        int cnt = 0;
        for(j=0;j<n;++j)
            {
            if(a[j] == 0)
                ++cnt;
            if(cnt == i + 1)
                break;
             
            dp[i][j] = Integer.MAX_VALUE;
            }
             
        for(;j<n;++j)
            {
            if(a[j] == 1)
                {
                dp[i][j] = dp[i][j-1];
                continue; 
                }    
                
            dp[i][j] = Math.abs(pos[i] - j) + dp[i-1][j-1];
            dp[i][j] = Math.min(dp[i][j], dp[i][j-1]); 
            }
        }  
    
//    for(i=0;i<one;++i)
//        {
//        for(j=0;j<n;++j)
//            output.write(dp[i][j] + " ");
//        output.write("\n");
//        }
    long ans = dp[one-1][n-1];
    output.write(ans + "\n");    
    output.flush();
    }
    public static void main(String[] args) throws Exception
    {
    int t = 1;
    //t = scan.nextInt();
    while(t-- > 0)
        tc();
    }
}
