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.
//package codeforces;
import java.io.PrintWriter;
import java.util.*;
public class codeforces {
	static int dp[][]=new int[5001][5001];
	public static void main(String[] args) {
		Scanner s=new Scanner(System.in);  
		PrintWriter out=new PrintWriter(System.out);
        int t=1;
        for(int tt=0;tt<t;tt++) {
        	int n=s.nextInt();
        	int a[]=new int[n];
        	ArrayList<Integer> z=new ArrayList<>();
        	ArrayList<Integer> o=new ArrayList<>();
        	for(int i=0;i<n;i++) {
        		a[i]=s.nextInt();
        		if(a[i]==1) {
        			o.add(i);
        		}else {
        			z.add(i);
        		}
        	}
        	for(int i=0;i<5001;i++) {
        		Arrays.fill(dp[i], -1);
        	}
        	System.out.println(sol(0,0,z,o));
        }
        out.close();
	    s.close();
	}
	
	static void sort(long[] a) {
		ArrayList<Long> l=new ArrayList<>();
		for (long i:a) l.add(i);
		Collections.sort(l);
		for (int i=0; i<a.length; i++) a[i]=l.get(i);
	}
	static void sortcol(int a[][],int c) {
		Arrays.sort(a, (x, y) -> {
			if (x[c] != y[c]) return(int)( x[c] - y[c]);
			return (int)-(x[1]+x[2] - y[1]-y[2]);
		});
	}
	static int gcd(int a, int b)
    {
      if (b == 0)
        return a;
      return gcd(b, a % b);
    }
	public static int sol(int i,int j,ArrayList<Integer> z,ArrayList<Integer> o) {
		if(j==o.size()) {
			return 0;
		}
		int h=z.size()-i;
		int l=o.size()-j;
		if(h<l) {
			return 10000000;
		}
		if(dp[i][j]!=-1) {
			return dp[i][j];
		}
		int ans1=sol(i+1,j,z,o);
		int ans2=sol(i+1,j+1,z,o)+Math.abs(z.get(i)-o.get(j));
		dp[i][j]=Math.min(ans1, ans2);
		return dp[i][j];
	}
}

