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.*;

public class D {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        ArrayList<Integer> occupied = new ArrayList<>();
        ArrayList<Integer> vacant = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            int x = scanner.nextInt();
            if (x == 1)
                occupied.add(i);
            else
                vacant.add(i);
        }

        Solution Solution = new Solution(occupied, vacant);

//        System.out.println(Solution.tabulation());
        System.out.println(Solution.memoization());
    }
}

class Solution {

    ArrayList<Integer> occupied, vacant;
    int x, y;

    public Solution(ArrayList<Integer> occupied, ArrayList<Integer> vacant) {
        this.occupied = occupied;
        this.vacant = vacant;
        x = occupied.size(); y = vacant.size();
    }

    int tabulation() {
        return tabulation(x, y);
    }

    int tabulation(int x, int y) {
        int[][] dp = new int[x+1][y+1];
        for (int i = 0; i <= x; i++) {
            Arrays.fill(dp[i], Integer.MAX_VALUE/2);
        }
        for (int i = 0; i <= x; i++) {
            dp[i][0] = 0;
        }
        for (int i = 0; i <= y; i++) {
            dp[0][i] = 0;
        }
        
        for (int i = 1; i <= x; i++) {
            for (int j = 1; j <= y; j++) {
                if(i == j) {
                    dp[i][j] = dp[i-1][j-1] + Math.abs(occupied.get(i-1) - vacant.get(j-1));
                }
                else {
                    dp[i][j] = Math.min(dp[i][j-1], dp[i-1][j-1] + Math.abs(occupied.get(i-1) - vacant.get(j-1)));
                }
            }
        }
        return dp[x][y];
    }

    int memoization() {
        int[][] dp = new int[x][y];
        for (int i = 0; i < x; i++) {
            Arrays.fill(dp[i], -1);
        }
        return memoization(dp, x-1, y-1);
    }

    int memoization(int[][] dp, int n, int m) {
        if(n < 0) {
            return 0;
        }
        if(m < n) {
            return Integer.MAX_VALUE;
        }
        if(dp[n][m] != -1) {
            return dp[n][m];
        }
        int first = memoization(dp, n, m-1);
        int second = memoization(dp, n-1, m-1) + Math.abs(occupied.get(n) - vacant.get(m));
        dp[n][m] = Math.min(first, second);
        return dp[n][m];
    }
}