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.*;
 public class ttrryy{
		public static void main (String[] args) throws java.lang.Exception
		{
			// your code goes here
			Scanner sc=new Scanner(System.in);
		   
			    int n=sc.nextInt();
			    int arr[]=new int[n];
			    int brr[]=new int[n];
			    int c=0;
			    for(int i=0;i<n;i++) {
			    	arr[i]=sc.nextInt();
			    	if(arr[i]==1) {
			    		c++;
			    	}
			    	brr[i]=(arr[i]^1);
			    }
			    int ans=0,ans1=0,min=5001,ind=-1;
			    HashSet<Integer> hs=new HashSet<>();
			   if(c>0) {
				   for(int i=0;i<n;i++) {
					   if(arr[i]!=1) {
						   continue;
					   }
					   min=5001;
					   ind=-1;
					   for(int j=0;j<n;j++) {
						   if(brr[j]==1 && !hs.contains(j)) {
							   if(Math.abs(j-i)<min) {
								   min=Math.abs(j-i);
								   ind = j;
							   }
						   }
					   }
					   ans+=min;
					   hs.add(ind);
				   }	
				   hs.clear();
				   for(int i=n-1;i>=0;i--) {
					   if(arr[i]!=1) {
						   continue;
					   }
					   min=5001;
					   ind=-1;
					   for(int j=0;j<n;j++) {
						   if(brr[j]==1 && !hs.contains(j)) {
							   if(Math.abs(j-i)<min) {
								   min=Math.abs(j-i);
								   ind = j;
							   }
						   }
					   }
					   ans1+=min;
					   hs.add(ind);
				   }

			   }
			   ans=Math.min(ans1, ans1);
			    System.out.println(ans);
			}
}