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 {

	static long m = (long) (1e9 + 7);

	public static void main(String[] args) throws IOException {
		Scanner scn = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		StringBuilder sb = new StringBuilder();
		int T = 1, tcs = 0;
		D: while (tcs++ < T) {
			int n = scn.ni();
			int a[] = scn.nia(n);
			ArrayList<Integer> l = new ArrayList<>();
			TreeSet<Integer> ts = new TreeSet<>();
			for (int i = 0; i < n; i++)
				if (a[i] == 1)
					l.add(i);
				else
					ts.add(i);
			int s = 0, e = l.size() - 1;
			boolean b = false;
			long ans = 0;
			while (s <= e) {
				int v;
				if (!b) {
					v = l.get(s);
					s++;
				} else {
					v = l.get(e);
					e--;
				}
				Integer f = ts.floor(v), c = ts.ceiling(v);
				if (f == null) {
					ts.remove(c);
					ans += c - v;
				} else if (c == null) {
					ts.remove(f);
					ans += v - f;
				} else {
					if (v - f < c - v) {
						ts.remove(f);
						ans += v - f;
					} else {
						ts.remove(c);
						ans += c - v;
					}
				}
				b = !b;
			}
			sb.append(ans);
			sb.append("\n");
		}
		out.print(sb);
		out.close();
	}

	static class Scanner {

		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public int ni() throws IOException {
			return Integer.parseInt(next());
		}

		public long nl() throws IOException {
			return Long.parseLong(next());
		}

		public int[] nia(int n) throws IOException {
			int a[] = new int[n];
			String sa[] = br.readLine().split(" ");
			for (int i = 0; i < n; i++)
				a[i] = Integer.parseInt(sa[i]);
			return a;
		}

		public long[] nla(int n) throws IOException {
			long a[] = new long[n];
			String sa[] = br.readLine().split(" ");
			for (int i = 0; i < n; i++)
				a[i] = Long.parseLong(sa[i]);
			return a;
		}

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

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