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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.StringTokenizer;

public class D {

  public static void main(String[] args) {
    // Write code here
    FastScanner fs = new FastScanner();
    int n = fs.nextInt();
    int[] state = fs.readArray(n);
    int k = Arrays.stream(state).sum();

    int[] pos = new int[k];
    for (int i = 0, j = 0; i < state.length; i++) {
      if (state[i] == 1) {
        pos[j] = i;
        j++;
      }
    }

    int[][] dp = new int[n + 1][k + 1];
    Arrays.stream(dp).forEach(row -> Arrays.fill(row, Integer.MAX_VALUE));
    dp[0][0] = 0;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j <= k; j++) {
        if (dp[i][j] == Integer.MAX_VALUE) {
          continue;
        }
        dp[i + 1][j] = Integer.min(dp[i + 1][j], dp[i][j]);
        if (j < k && state[i] == 0) {
          dp[i + 1][j+1] = Integer.min(dp[i + 1][j+1], dp[i][j] + Math.abs(pos[j] - i));
        }
      }
    }

    System.out.println(dp[n][k]);
  }

  static void sort(int[] a) {
    ArrayList<Integer> l = new ArrayList<>();
    for (int i : a) {
      l.add(i);
    }
    Collections.sort(l);
    for (int i = 0; i < a.length; i++) {
      a[i] = l.get(i);
    }
  }

  static class FastScanner {

    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st = new StringTokenizer("");

    String next() {
      while (!st.hasMoreTokens()) {
        try {
          st = new StringTokenizer(br.readLine());
        } catch (IOException e) {
          e.printStackTrace();
        }
      }
      return st.nextToken();
    }

    int nextInt() {
      return Integer.parseInt(next());
    }

    int[] readArray(int n) {
      int[] a = new int[n];
      for (int i = 0; i < n; i++) {
        a[i] = nextInt();
      }
      return a;
    }

    long nextLong() {
      return Long.parseLong(next());
    }
  }
}
