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.*;
import java.util.*;

public class D {
	
	static int[][] dp;
	
	static int solve(int i,int j,ArrayList<Integer> A,ArrayList<Integer> B) {
		if(i==A.size()) {
			return 0;
		}
		if(j==B.size()) {
			return 1000000009;
		}
		if(dp[i][j]!=-1) {
			return dp[i][j];
		}
		int ans=1000000009;
		int a=A.get(i);
		int b=B.get(j);
		ans=Math.min(ans, Math.abs(a-b)+solve(i+1,j+1,A,B));
		ans=Math.min(ans, solve(i,j+1,A,B));
		return dp[i][j]=ans;
	}

	public static void main(String[] args){
		FastReader sc=new FastReader();
		int t=1;
		while(t-->0) {
			int n=sc.nextInt();
			int[] a=new int[n];
			ArrayList<Integer> A=new ArrayList<>();
			ArrayList<Integer> B=new ArrayList<>();
			for(int i=0;i<n;i++) {
				a[i]=sc.nextInt();
				if(a[i]==1) {
					A.add(i);
				} else {
					B.add(i);
				}
			}
			dp=new int[5010][5010];
			for(int i=0;i<5010;i++) {
				Arrays.fill(dp[i], -1);
			}
			System.out.println(solve(0,0,A,B));
			
		}
	}
	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader()
		{
			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;
		}
	}

}
