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 B{
	static int n;
	static int m;
	static long[][] dp;
	static int[] pos;
	static int[] emp;
	
	static long make(int i, int j)
	{
		if(i>=n)return 0;
		if(j>=m)return (long)(1e12);
		if(dp[i][j] != -1)return dp[i][j];
		long ans = Math.abs(pos[i]-emp[j]) + make(i+1,j+1);
		ans = Math.min(ans,make(i,j+1));
		return dp[i][j] = ans;
	}
	
	public static void main(String[] args)
	{
		FastScanner fs = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		n = fs.nextInt();
		List<Integer> temp = new ArrayList();
		List<Integer> pip = new ArrayList();
		int[] arr = new int[n];
		for(int i=0;i<n;i++)
		{
			arr[i] = fs.nextInt();
			if(arr[i] == 0)
			{
				temp.add(i);
			}
			else
			{
				pip.add(i);
			}
		}
		n = pip.size(); m = temp.size();
		int start = 0;
		pos = new int[pip.size()];
		for(int i:pip)pos[start++] = i;
		start = 0;
		emp = new int[temp.size()];
		for(int i:temp)emp[start++] = i;
		dp = new long[n][m];
		for(int i=0;i<n;i++)for(int j=0;j<m;j++)dp[i][j] = -1;
		out.println(make(0,0));
		out.close();
	}
	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());
		}
	}
	public static int[] sort(int[] arr)
	{
		List<Integer> temp = new ArrayList();
		for(int i:arr)temp.add(i);
		Collections.sort(temp,Collections.reverseOrder());
		int start = 0;
		for(int i:temp)arr[start++]=i;
		return arr;
	}
	public static char[] rev(char[] arr)
	{
		char[] ret = new char[arr.length];
		int start = arr.length-1;
		for(char i:arr)ret[start--] = i;
		return ret;
	}
}