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 {
	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];
        	for(int i=0;i<n;i++) {
        		a[i]=s.nextInt();
        	}
        	int dp[][]=new int[5050][10100];
        	for(int i=0;i<5050;i++) {
        		for(int j=0;j<10100;j++) {
        			dp[i][j]=10000000;
        		}
        	}
        	dp[0][5050]=0;
        	for(int i=1;i<=n;i++){
                if(a[i-1] == 0){
                    for(int j = -i+5050;j<5050;j++){
                        dp[i][j] = Math.min(dp[i-1][j],dp[i-1][j+1]-i);
                    }
                    for(int j = 5050;j<=i+5050;j++){
                        dp[i][j] =Math. min(dp[i-1][j],dp[i-1][j+1]+i);
                    }
                }else{
                    for(int j = -i+5050;j<=5050;j++){
                        dp[i][j] = dp[i-1][j-1]+i;
                    }
                    for(int j = 5051;j<=i+5050;j++){
                        dp[i][j] = dp[i-1][j-1]-i;
                    }
                }
            }
        	System.out.println(dp[n][5050]);
        }
        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);
    }
	
}

