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.lang.*;
import java.io.*;
public class Main
{
    static long dp[][];
    static long recurse(int arr0[],int arr1[],int i,int j){
        if(i==arr1.length) return 0;
        if(j==arr0.length) return 100000000000l;
        if(dp[i][j]>=0) return dp[i][j];
        long min;
        //for(int i=p;i<arr1.length;i++){
            min=Math.min(recurse(arr0,arr1,i+1,j+1)+Math.abs(arr1[i]-arr0[j]),
            recurse(arr0,arr1,i,j+1));
       // }
        return dp[i][j]=min;
    }
    
	public static void main (String[] args) throws java.lang.Exception
	{
	    Scanner sc=new Scanner(System.in);
	   // BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	    int t=1;
	   // t=sc.nextInt();
	    //int t=Integer.parseInt(br.readLine());
	    while(--t>=0){
	        int n=sc.nextInt();
	        dp=new long[n+5][n+5];
	        int occ=0,unocc=0;
	        int a[]=new int[n];
	        for(int i=0;i<n;i++){
	            a[i]=sc.nextInt();
	            if(a[i]==0) unocc++;
	            else occ++;
	        }
	        for(int i=0;i<n+5;i++)for(int j=0;j<n+5;j++) dp[i][j]=-1;
	        int arr0[]=new int[unocc];
	        int arr1[]=new int[occ];
	        unocc=0;occ=0;
	        for(int i=0;i<n;i++){
	            if(a[i]==1)arr1[occ++]=i;
	            else arr0[unocc++]=i;
	        }
	       // for(int i=0;i<occ;i++)System.out.println(arr1[i]);
	           System.out.println(recurse(arr0,arr1,0,0));
	        
	    }
	    
	}
}