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 Soln {
	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

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

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

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

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}
	public static void main(String[] args) {
		try {
			System.setIn(new FileInputStream("input.txt"));
			System.setOut(new PrintStream(new FileOutputStream("output.txt")));
		} catch (Exception e) {
			System.err.println("Error");
		}

		FastReader fs = new FastReader();
		int t = 1;
		while(t-- > 0) {
			int n = fs.nextInt();
			int[] a = new int[n];
			for(int i=0; i<n; i++) {
				a[i] = fs.nextInt();
			}
			int ans = 0;

			boolean[] visited = new boolean[n];
			for(int i=0; i<n; i++) {
				if(a[i] == 1 && !visited[i]) {
					int l = i-1;
					int h = i+1;
					int inx = -1;
					while(l >= 0) {
						if(a[l] == 0 && !visited[l]) {
							break;
						}
						l--;
					}
					while(h < n) {
						if(a[h] == 0 && !visited[h]) {
							break;
						}
						h++;
					}

					int left = (l != -1) ? i-l : Integer.MAX_VALUE;
					int right = (h != n) ? h-i : Integer.MAX_VALUE;

					if(left < right) {
						inx = l;
						ans += left;
					} else {
						inx = h;
						ans += right;
					}

					int temp = a[i];
					a[i] = a[inx];
					a[inx] = temp;

					visited[inx] = true;
					visited[i] = true;
					
				}
				
			}
			System.out.println(ans);
		}

			
	}
}