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.lang.*;
import java.io.*;

public class Main {
    static int func(List<Integer> ones, List<Integer> zeroes, int idO, int idZ, int o, int z, Integer[][] memo) {
        if(idZ == zeroes.size() || idO == ones.size()) {
            return 0;
        }
        
        if(memo[idO][idZ] != null) {
            return memo[idO][idZ];
        }
        
        if(o == z) {
            return memo[idO][idZ] = Math.abs(ones.get(idO) - zeroes.get(idZ))
                + func(ones, zeroes, idO + 1, idZ + 1, o - 1, z - 1, memo);
        } else {
            return memo[idO][idZ] = Math.min(
                Math.abs(ones.get(idO) - zeroes.get(idZ))
                + func(ones, zeroes, idO + 1, idZ + 1, o - 1, z - 1, memo),
                func(ones, zeroes, idO, idZ + 1, o, z - 1, memo)
            );
        }
    }
    
	public static void main (String[] args) {
	   // Scanner scan = new Scanner(System.in);
    	FastScanner scan = new FastScanner();
        int n =  scan.nextInt();
        int[] arr = scan.readArray(n);
        
        List<Integer> ones = new ArrayList<>();
        List<Integer> zeroes = new ArrayList<>();
        
        for(int i = 0; i < n; i++) {
            if(arr[i] == 0) {
                zeroes.add(i);
            } else {
                ones.add(i);
            }
        }
        
        int o = ones.size();
        int z = zeroes.size();
        
        Integer[][] memo = new Integer[o + 1][z + 1];
        
        int ans = func(ones, zeroes, 0, 0, o, z, memo);
        
        System.out.println(ans);
    }
/*
    int n = scan.nextInt();
            int[] arr = scan.readArray(n);
    
    int n = scan.nextInt();
            int m = scan.nextInt();
            int[][] arr = new int[n][m];
            for(int i = 0; i < n; i++) {
                for(int j = 0; j < m; j++) {
                    arr[i][j] = scan.nextInt();
                }
            }
    
*/
    
    
    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());
		}
	}
    
}