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 D {

	public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	public static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));

	public static void main(String[] args) throws IOException {
		readInput();
		out.close();
	}
	
	static List<Integer> o1, o0;
	static int[][] dp;
	
	static int solve(int i, int j) {
		if (i >= dp.length) return 0;
		if (j >= dp[0].length) return Integer.MAX_VALUE/10;
		
		if (dp[i][j] == -1) {
			dp[i][j] = Integer.min(solve(i,j+1), solve(i+1,j+1) + Math.abs(o1.get(i)-o0.get(j)));
		}
		return dp[i][j];
	}
	
	public static void readInput() throws IOException {
		// br = new BufferedReader(new FileReader(".in"));
		// out = new PrintWriter(new FileWriter(".out"));
		int n;
		n = Integer.parseInt(br.readLine());
		int[] a=  new int[n];
		StringTokenizer st = new StringTokenizer(br.readLine());
		o1 = new ArrayList<Integer>();
		o0 = new ArrayList<Integer>();
		for (int i = 0; i < n; i++) {
			a[i] = Integer.parseInt(st.nextToken());
			if (a[i] == 1) o1.add(i);
			else o0.add(i);
		}
		if (o1.size() == 0) {
			out.println(0);
			return;
		}
		dp = new int[o1.size()][o0.size()];
		for (int[] x: dp) Arrays.fill(x, -1);
		out.println(solve(0,0));
		
	}
}
