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.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.PriorityQueue;
import java.util.StringTokenizer;
import java.io.*;
 
public class Armchairs {
	
	private static class MyScanner {
	    BufferedReader br;
	    StringTokenizer st;

	    public MyScanner() {
	       br = new BufferedReader(new InputStreamReader(System.in));
	    }

	    String next() {
	        while (st == null || !st.hasMoreElements()) {
	            try {
	                st = new StringTokenizer(br.readLine());
	            } catch (IOException e) {
	                e.printStackTrace();
	            }
	        }
	        return st.nextToken();
	    }

	    int nextInt() {
	        return Integer.parseInt(next());
	    }

	    long nextLong() {
	        return Long.parseLong(next());
	    }

	    double nextDouble() {
	        return Double.parseDouble(next());
	    }

	    String nextLine(){
	        String str = "";
		  try {
		     str = br.readLine();
		  } catch (IOException e) {
		     e.printStackTrace();
		  }
		  return str;
	    }

	 }
	

	public static long[][] dp; 
	
	public static long rec(int n,  int c, HashMap<Integer, Integer> pos, int[] arr)
	
	{
		
		if(dp[n][c]!=-2)
			return dp[n][c];
		
		if(c==0)
			{ dp[n][c] = 0; return 0; }
		
		if(n==0)
		   { dp[n][c] = -1; return -1; }
			  
		if(arr[n-1]==1)
		   dp[n][c] = rec(n-1,c,pos,arr);
		
		if(arr[n-1]==0)
		   { 
		     long x = rec(n-1,c,pos,arr);
		     long y = rec(n-1,c-1,pos,arr);
		     
		     if(x==-1 && y!=-1)
		    	dp[n][c] = y + Math.abs(pos.get(c)-n+1); 
		     
		     if(x!=-1 && y==-1)
			    	dp[n][c] = x;
		     
		     if(x==-1 && y==-1)
			    	dp[n][c] = -1; 
		     
		     if(x!=-1 && y!=-1)
			    	dp[n][c] = Math.min(x, y + Math.abs(pos.get(c)-n+1)) ; 
		    	 
		   }
		
	    return dp[n][c];
	}
	
	public static long solution(int[] arr, int n)
			
	{
		
		int one = 0;
		HashMap<Integer, Integer> pos = new HashMap<>();
		
		for(int i = 0; i<n; i++)
		{
			if(arr[i]==1)
				{ one++; pos.put(one, i); }
		}
		
		dp = new long[n+1][one+1];
		
		for(int i = 0; i<=n; i++)
		{
			for(int j = 0; j<=one; j++)
				dp[i][j] = -2;
		}
		
		return  rec(n,one,pos,arr);
		
	}
        
private static PrintWriter out = new PrintWriter(System.out);

public static void main (String[] args)
{
	MyScanner s =  new MyScanner();
     
    int n = s.nextInt();
    int[] arr = new int[n];
    for(int i =0; i<n; i++)
      arr[i] = s.nextInt();
    	
    out.println(solution(arr,n));
    
    
    out.flush();
    out.close();
    
}
}
