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 E {

	public static void main(String[] args) throws NumberFormatException, IOException {
		// TODO Auto-generated method stub

		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		
		int n = Integer.parseInt(br.readLine());
		int cnt = n;
		boolean[] non = new boolean[n];
		StringTokenizer st = new StringTokenizer(br.readLine());
		
		for(int i = 0; i < n; i++) {
			
			if(Integer.parseInt(st.nextToken()) == 0) {
				non[i] = true;
				cnt--;
			}
			
		}
		
		int x = 0;
		int y = 0;
		int[] location = new int[cnt];
		int[] rlocation = new int[n-cnt];
		
		for(int i = 0; i < n; i++) {
			
			if(!non[i]) {
				
				location[x] = i;
				x++;
				
			}else{
				
				rlocation[y] = i;
				y++;
				
			}
			
		}
		
		int[][] dp = new int[(n-cnt)+1][cnt+1];
		Arrays.fill(dp[0], 100000000);
		dp[0][0] = 0;
		
		for(int i = 0; i < n-cnt; i++) {
			
			//System.out.println("HIT");
			
			if(i < (n-cnt))
				Arrays.fill(dp[i+1], 100000000);
						
			for(int j = 0; j < cnt; j++) {
				
				if(i < (n-cnt)) {
					dp[i+1][j] = Math.min(dp[i+1][j], dp[i][j]);
					dp[i+1][j+1] = Math.min(dp[i+1][j+1], dp[i][j] + Math.abs(rlocation[i] - location[j]));			
					//System.out.println(dp[i+1][j+1] + " " + dp[i][j] + " " + j + " " + rlocation[i] + " " + location[j]);
				}
				
			}
			
		}
		
		int min = Integer.MAX_VALUE;
		
		for(int i = 0; i < (n-cnt)+1; i++) {
			
			min = Math.min(dp[i][cnt], min);
			
		}
				
		System.out.println(min);
		
	}

}