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.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.PriorityQueue;
import java.util.Scanner;

public class Armchairs {

	static List<Integer> zpos ;
	static List<Integer> opos ;
	
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		zpos = new ArrayList<Integer>();
		opos = new ArrayList<Integer>();
		for (int i = 0; i < n; i++) {
			int x = sc.nextInt();
			if (x == 1) opos.add(i);
			else zpos.add(i);
		}
		long[][] dp = new long[5001][5001];
		for (int i = 0; i < 5001; i++) Arrays.fill(dp[i], -1);
		System.out.println(dp(0, 0, opos.size(), zpos.size(), opos.size(), dp));
	}

	private static long dp(int i, int j, int m, int n, int cnt, long[][] dp) {
		// TODO Auto-generated method stub
		if (cnt == 0) return 0;
		if (dp[i][j] != -1) return dp[i][j];
		if (i >= m || j >= n) return Integer.MAX_VALUE;
		long dns = dp(i, j + 1, m, n, cnt, dp);
		long s = Math.abs(zpos.get(j) - opos.get(i)) + dp(i + 1, j + 1, m, n, cnt - 1, dp);
		return dp[i][j] = Math.min(dns, s);
	}

}
