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.*;
/**
 * Created by Brandon on 4/29/2021, 11:04 AM.
 */
public class Main {
    static long mod = 998244353;

    public static void solve() {
        // string:
        //char** vals = new char* [n];
        //for (int i = 0; i < n; i++) {
        //	string str = "";
        //	cin >> str;
        //	vals[i] = new char[n];
        //	for (int j = 0; j < n; j++) {
        //		vals[i][j] = str[j];
        //	}
        //}
        int n;
        Scanner sc = new Scanner(System.in);
        n = Integer.parseInt(sc.next());
	    int[] arr = new int[n];
        ArrayList<Integer> ones = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            arr[i] = Integer.parseInt(sc.next());
            if (arr[i] == 1) {
                ones.add(i);
            }
        }
        if (ones.size() == 0) {
            System.out.println("0\n");
            return;
        }

	    int[] zeroes = new int[n - ones.size()];
        int cur = 0;
        for (int i = 0; i < n; i++) {
            if (arr[i] == 0) {
                zeroes[cur] = i;
                cur++;
            }
        }
	    int[][] dp = new int[ones.size()][n-ones.size()];
        for (int i = 1; i < ones.size(); i++) {
            dp[i][0] = 100000000;
        }
        dp[0][0] = Math.abs(ones.get(0) - zeroes[0]);
        for (int i = 1; i < n - ones.size(); i++) {
            dp[0][i] = Math.min(dp[0][i - 1], Math.abs(ones.get(0) - zeroes[i]));
        }
        // dp[i][j] is number where you only have to move first i but max position you can put is j
        for (int i = 1; i < ones.size(); i++) {
            for (int j = 1; j < n - ones.size(); j++) {
                dp[i][j] = Math.min(dp[i][j - 1], dp[i-1][j-1] + Math.abs(zeroes[j] - ones.get(i)));
            }
        }
        System.out.println(dp[ones.size()-1][n-ones.size()-1] + "\n");
    }
    public static void main(String[] args) {
        solve();
    }
}
