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 Main{
        public static void main(String[] args){
        	FastScanner in=new FastScanner(System.in);
        	
        	int t=in.nextInt();
        	PriorityQueue<Integer> person = new PriorityQueue<>();
        	
        	PriorityQueue<Integer> set = new PriorityQueue<>(new Comparator<Integer>() {

        		@Override
        			public int compare(Integer o1, Integer o2) {
        			// TODO Auto-generated method stub
        				return o2-o1;
        			}
        	        		
        	        		
        	});
        	int ans=0;
        	for(int i=0;i<t;i++) {
        		if(in.nextInt()==1) {
        			if(set.isEmpty()) {
        				person.add(i);
        			}else {
        				
        				ans=ans+i-set.poll();
        			}
        		}else {
        			if(person.isEmpty()) {
        				set.add(i);
        			}else {
        				ans=ans+i-person.poll();
        			}
        		}
        	}
        	System.out.print(ans);
        }
        
        
        
        static class FastScanner{//用于快速读入大量数据
    		BufferedReader br;
    		StringTokenizer st;
    		public FastScanner(InputStream in) {
    			br = new BufferedReader(new InputStreamReader(in),16384);
    			eat("");
    		}
    		public void eat(String s) {
    			st = new StringTokenizer(s);
    		}
    		public String nextLine() {
    			try {
    				return br.readLine();
    			} catch (IOException e) {
    				return null;
    			}
    		}
    		public boolean hasNext() {
    			while(!st.hasMoreTokens()) {
    				String s = nextLine();
    				if(s==null) return false;
    				eat(s);
    			}
    			return true;
    		}
    		public String next() {
    			hasNext();
    			return st.nextToken();
    		}
    		public int nextInt() {
    			return Integer.parseInt(next());
    		}
    		public long nextLong() {
    			return Long.parseLong(next());
    		}
    	}
    }