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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class minMax {

	public static void main(String[] args) {
		FastReader sc = new FastReader();
		int n = sc.nextInt();

		int[] arr = new int[n];
		int[] nums = new int[n];

		for (int i = 0; i < n; i++) {
			int a = sc.nextInt();
			arr[i] = a;
			nums[i] = a;
		}

		int time1 = 0;
		int time2 = 0;

		for (int i = 0; i < n; i++) {
			if (arr[i] == 1) {
				int j = i - 1;
				int k = i + 1;
				for (j = i - 1; j >= 0; j--) {
					if (arr[j] == 0)
						break;
				}
				for (k = i + 1; k < n; k++) {
					if (arr[k] == 0)
						break;
				}

				int l = Math.abs(i - j);
				int r = Math.abs(k - i);
				if ((l < r && j != -1) || k == n) {
					arr[j] = -1;
					time1 += l;
				} else if ((r < l && k != n) || j == -1) {
					arr[k] = -1;
					time1 += r;
				} else if (l == r) {
					time1 += l;
					if (l == -1) {
						arr[k] = -1;
					} else {
						arr[j] = -1;
					}
				}
			}
		}

		for (int i = n - 1; i >= 0; i--) {
			if (nums[i] == 1) {
				int j = i - 1;
				int k = i + 1;
				for (j = i - 1; j >= 0; j--) {
					if (nums[j] == 0)
						break;
				}
				for (k = i + 1; k < n; k++) {
					if (nums[k] == 0)
						break;
				}

				int l = Math.abs(i - j);
				int r = Math.abs(k - i);
				if ((l < r && j != -1) || k == n) {
					nums[j] = -1;
					time2 += l;
				} else if ((r < l && k != n) || j == -1) {
					nums[k] = -1;
					time2 += r;
				} else if (l == r) {
					time2 += l;
					if (l == -1) {
						nums[k] = -1;
					} else {
						nums[j] = -1;
					}
				}
			}
		}

		System.out.println(Math.min(time1, time2));

	}

	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;
		}
	}
}
