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.io.*;
public class Armchairs 
{
	public static int val(int b[],int m,int c[],int n,int dp[][])
	{
		if(dp[m][n]>-1)
		{
			return dp[m][n];
		}
		else
		if(m==0)
		{
			return 0;
		}
		else
		if(n==0)
		{
							return 2099999999;
		}
		else
		{
		
			return dp[m][n]=Math.min(val(b,m-1,c,n-1,dp)+Math.abs(b[m-1]-c[n-1]),val(b,m,c,n-1,dp));
		}
	}
	public static void process()throws IOException
	{
		int n=I();
		int a[]=Ai(n);
		ArrayList<Integer> arr=new ArrayList<Integer>();
		ArrayList<Integer> arr1=new ArrayList<Integer>();
		for(int i=0;i<n;i++)
		{
			if(a[i]==0)
			{
				arr1.add(i);
			}
			else
			{
				arr.add(i);
			}
		}
		int b[]=new int[arr.size()];
		int c[]=new int[arr1.size()];
		int m=arr.size();int n1=arr1.size();
		int dp[][]=new int[m+1][n1+1];

		dyn(dp,m+1,n1+1,-1);
		
		for(int i=0;i<arr.size();i++)
		{
			b[i]=arr.get(i);
		}
		for(int i=0;i<arr1.size();i++)
		{
			c[i]=arr1.get(i);
		}

		arr.clear();
	    arr1.clear();
	   pn(val(b,m,c,n1,dp));
	}
	static Scanner sc = new Scanner(System.in);
	static PrintWriter out = new PrintWriter(System.out);
	static void pn(Object o){out.println(o);out.flush();}
	static void p(Object o){out.print(o);out.flush();}
	static void pni(Object o){out.println(o);System.out.flush();}
	static int I() throws IOException{return sc.nextInt();}
	static long L() throws IOException{return sc.nextLong();}
	static double D() throws IOException{return sc.nextDouble();}
	static String S() throws IOException{return sc.next();}
	static char C() throws IOException{return sc.next().charAt(0);}
	static int[] Ai(int n) throws IOException{int[] arr = new int[n];for (int i = 0; i < n; i++)arr[i] = I();return arr;}
	static String[] As(int n) throws IOException{String s[] = new String[n];for (int i = 0; i < n; i++)s[i] = S();return s;}
	static long[] Al(int n) throws IOException {long[] arr = new long[n];for (int i = 0; i < n; i++)arr[i] = L();return arr;}
	static void dyn(int dp[][],int n,int m,int z)throws IOException {for(int i=0;i<n;i++){ for(int j=0;j<m;j++){ dp[i][j]=z;}} }
	
//	*--------------------------------------------------------------------------------------------------------------------------------*//
	static class AnotherReader {BufferedReader br;StringTokenizer st;AnotherReader() throws FileNotFoundException {br = new BufferedReader(new InputStreamReader(System.in));}
	AnotherReader(int a) throws FileNotFoundException {br = new BufferedReader(new FileReader("input.txt"));}
	String next() throws IOException{while (st == null || !st.hasMoreElements()) {try {st = new StringTokenizer(br.readLine());} catch (IOException e) {e.printStackTrace();}}return st.nextToken();}
	int nextInt() throws IOException {return Integer.parseInt(next());}
	long nextLong() throws IOException {return Long.parseLong(next());}
	double nextDouble() throws IOException {return Double.parseDouble(next());}
	String nextLine() throws IOException {String str = "";try {str = br.readLine();} catch (IOException e) {e.printStackTrace();}return str;}}
	public static void main(String[] args)throws IOException{try{boolean oj=true;if(oj==true)
	{AnotherReader sk=new AnotherReader();PrintWriter out=new PrintWriter(System.out);}
	else
	{AnotherReader sk=new AnotherReader(100);out=new PrintWriter("output.txt");}
	//long T=L();while(T-->0)
	{process();}out.flush();out.close();}catch(Exception e){return;}}}
//*-----------------------------------------------------------------------------------------------------------------------------------*//
	
