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.*;

public class Binary_search {
	public static int get(List<Integer> vals,int key) {
		int lo=0;
		int hi=vals.size()-1;
		int x=rank(vals,lo,hi,key);
		return x;
	}
	public static int rank(List<Integer> vals,int lo,int hi,int key){
		while(lo<hi) {
			int mid=lo+(hi-lo)/2;
			if(key<vals.get(mid)) hi=mid-1;
			else if(key>vals.get(mid)) lo=mid+1;
			else if(key==vals.get(mid)) return mid;
		}
		return lo;
	}
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		List<Integer> ones = new LinkedList<>();
		List<Integer> zeros = new LinkedList<>();
		int cost = 0;
		for(int i=0;i<n;i++) {
			int x = sc.nextInt();
			if(x==0) {
				zeros.add(i+1);
			}
			else {
				ones.add(i+1);
			}
		}
		for(int i=0;i<ones.size();i++) {
			int num = ones.get(i);
			int x = get(zeros, num);
			if(zeros.get(x)>ones.get(i)) {
				if(x-1>=0) {
					if(zeros.get(x)-ones.get(i)<Math.abs(zeros.get(x-1)-ones.get(i))) {
						cost+=zeros.get(x)-ones.get(i);
						zeros.remove(x);
					}
					else {
						cost+=Math.abs(zeros.get(x-1)-ones.get(i));
						zeros.remove(x-1);
					}
				}
				else {
					cost+=zeros.get(x)-ones.get(i);
					zeros.remove(x);
				}
			}
			else {
				if(x+1<zeros.size()) {
					if(Math.abs(zeros.get(x)-ones.get(i))<=zeros.get(x+1)-ones.get(i)) {
						cost+=Math.abs(zeros.get(x)-ones.get(i));
						zeros.remove(x);
					}
					else {
						cost+=zeros.get(x+1)-ones.get(i);
						zeros.remove(x+1);
					}
				}
				else {
					cost+=Math.abs(zeros.get(x)-ones.get(i));
					zeros.remove(x);
				}
			}
		}
		System.out.println(cost);
    }
}