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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
 
 
 
public class B {
	static int mod=1000_000_007;
	
	public static void main(String[] args) throws Exception {
		PrintWriter out=new PrintWriter(System.out);
	    FastScanner fs=new FastScanner();
	    int n=fs.nextInt();
	    int arr[]=fs.readArray(n);
	    TreeSet<Integer> tree=new TreeSet<>();
	    for(int i=0;i<n;i++) if(arr[i]==0) tree.add(i);
	    int ans=Integer.MAX_VALUE;
	    for(int j=0;j<n;j++) {
	    	TreeSet<Integer> set=new TreeSet<>(tree);
	    	int sum=0;
	    	for(int i=0;i<=j;i++) {
	    		if(arr[i]==1) {
		    		Integer floor=set.floor(i);
		    		if(floor!=null) {
		    			sum+=(i-floor);
		    			set.remove(floor);
		    		}
		    		else {
		    			Integer ceil=set.ceiling(i);
		    			sum+=(ceil-i);
		    			set.remove(ceil);
		    		}
		    	}
	    	}
	    	for(int i=j+1;i<n;i++) {
	    		if(arr[i]==1) {
	    			Integer ceil=set.ceiling(i);
		    		Integer floor=set.floor(i);
		    		if(ceil!=null) {
		    			sum+=(ceil-i);
		    			set.remove(ceil);
		    		}
		    		else {
		    			
		    			sum+=(i-floor);
		    			set.remove(floor);
		    		}
		    	}
	    	}
	    	ans=Math.min(ans, sum);
	    }
//	    for(int i=0;i<n;i++) {
//	    	if(arr[i]==1) {
//	    		Integer floor=set.floor(i);
//	    		Integer ceil=set.ceiling(i);
//	    		if(floor!=null&&ceil!=null) {
//	    			if(i-floor<=ceil-i) {
//	    				sum+=i-floor;
//	    				set.remove(floor);
//	    			}
//	    			else {
//	    				sum+=ceil-i;
//	    				set.remove(ceil);
//	    			}
//	    		}
//	    		else if(floor!=null) {
//	    			sum+=i-floor;
//    				set.remove(floor);
//	    		}
//	    		else {
//	    			sum+=ceil-i;
//    				set.remove(ceil);
//	    		}
//	    	}
//	    }
//	    int ans=0;
//	    for(int i=0;i<n;i++) {
//	    	if(arr[i]==1) {
//	    		Integer floor=set.floor(i);
//	    		Integer ceil=set.ceiling(i);
//	    		if(floor!=null&&ceil!=null) {
//	    			if(i-floor>=ceil-i) {
//	    				ans+=ceil-i;
//	    				set.remove(ceil);
//	    				
//	    			}
//	    			else {
//	    				ans+=i-floor;
//	    				set.remove(floor);
//	    			}
//	    		}
//	    		else if(floor!=null) {
//	    			ans+=i-floor;
//    				set.remove(floor);
//	    		}
//	    		else {
//	    			ans+=ceil-i;
//    				set.remove(ceil);
//	    		}
//	    	}
//	    }
//	     sum=Math.min(sum, ans);
	    System.out.println(ans);
	}
	
	static void sort(int[] a) {
		//suffle
		int n=a.length;
		Random r=new Random();
		for (int i=0; i<a.length; i++) {
			int oi=r.nextInt(n);
			int temp=a[i];
			a[i]=a[oi];
			a[oi]=temp;
		}
		
		//then sort
		Arrays.sort(a);
	}
	
	// Use this to input code since it is faster than a Scanner
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
 
}