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.Scanner;

public class Armchair {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scanner=new Scanner(System.in);
		int n=scanner.nextInt();
		int arr[]=new int[n];
		for(int i=0;i<n;i++)
		{
			arr[i]=scanner.nextInt();
		}
		int count=0;
		for(int i=0;i<n;i++)
		{
			if(arr[i]==1)
			{
				int flag_=0;
				if(!(i-1<0))
				{
					if(arr[i-1]==0|| arr[i+1]==0)
					{
						count+=1;
						if(arr[i-1]==0)
							arr[i-1]=2;
						else 
							arr[i+1]=2;
						flag_=1;
					}
				}
				else if (arr[i+1]==0)
				{
					count+=1;
					flag_=1;
					arr[i+1]=2;
				}
				
					if(flag_==0)
					{
						int j=i+1;
						int flag=0;
						while (j<arr.length)
						{
							if(arr[j]==0)
							{
								arr[j]=2;
								count+=j-i;
								flag=1;
								break;
							}
							j++;
						}
						if(flag==0)
						{
							j=i-1;
							while(j>=0)
							{
								if(arr[j]==0)
								{
									arr[j]=2;
									count+=i-j;
									flag=1;
									break;
								}
								j--;
							}
						}
					}
				//check if th
				//find the 1st position which is non zero and then set it to 2;
				
			}
		}
		System.out.println(count);
	}

}
