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 q3 {

    public static void main(String[] args) throws Exception {
        // String[] parts=br.readLine().split(" ");
        // int n=Integer.parseInt(parts[0]);
        // int k=Integer.parseInt(parts[1]);
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        // int tests = Integer.parseInt(br.readLine());

        int n = Integer.parseInt(br.readLine());
        String[] parts=br.readLine().split(" ");

        int[] arr = new int[n];
        for(int i = 0;i < parts.length;i++) arr[i] = Integer.parseInt(parts[i]);

        int[] cones = new int[n];
        int[] czeroes = new int[n];
        int co = 0,cz = 0;
        for(int i = n - 1;i >= 0;i--){
            cones[i] = co;
            czeroes[i] = cz;
            if(arr[i] == 1) co++;
            else cz++;
        }

        int ans = 0;
        for(int i = 0;i < n;i++){
            if(arr[i] == 1){
                if(cones[i] >= czeroes[i]){
                    int l = i - 1;
                    while(l >= 0 && arr[l] != 0) l--;
                    arr[l] = -1;
                    ans += i - l;
                }else{
                    int tans = Integer.MAX_VALUE;

                    int l = i - 1;
                    while(l >= 0 && arr[l] != 0) l--;
                    if(l >= 0) tans = i - l;

                    int r = i + 1;
                    while(r < n && arr[r] != 0) r++;
                    if(r < n && r - i < tans) {
                        ans += r - i;

                        for(int j = r-1;j >i;j--) czeroes[j]--;
                        arr[r] = -1;
                    }else{
                        ans += i - l;
                        arr[l] = -1;
                    }

                }
                
            }
        }
        
        System.out.println(ans);
    }
}
