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.*;


public class Main {
    static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            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;
        }

    }
    static class TreeMultiSet<T> implements Iterable<T>
    {
        private final TreeMap<T,Integer> map;
        private int size;

        public TreeMultiSet(){map=new TreeMap<>(); size=0;}
        public TreeMultiSet(boolean reverse)
        {
            if(reverse) map=new TreeMap<>(Collections.reverseOrder());
            else map=new TreeMap<>(); size=0;
        }
        public void clear(){map.clear(); size=0;}
        public int size(){return size;}
        public int setSize(){return map.size();}
        public boolean contains(T a){return map.containsKey(a);}
        public boolean isEmpty(){return size==0;}
        public Integer get(T a){return map.getOrDefault(a,0);}
        public void add(T a, int count)
        {
            int cur=get(a);map.put(a,cur+count); size+=count;
            if(cur+count==0) map.remove(a);
        }
        public void addOne(T a){add(a,1);}
        public void remove(T a, int count){add(a,Math.max(-get(a),-count));}
        public void removeOne(T a){remove(a,1);}
        public void removeAll(T a){remove(a,Integer.MAX_VALUE-10);}

        public T ceiling(T a){return map.ceilingKey(a);}
        public T floor(T a){return map.floorKey(a);}
        public T first(){return map.firstKey();}
        public T last(){return map.lastKey();}
        public T higher(T a){return map.higherKey(a);}
        public T lower(T a){return map.lowerKey(a);}
        public T pollFirst(){T a=first(); removeOne(a); return a;}
        public T pollLast(){T a=last(); removeOne(a); return a;}

        public Iterator<T> iterator()
        {
            return new Iterator<>()
            {
                private final Iterator<T> iter = map.keySet().iterator();
                private int count = 0; private T curElement;

                public boolean hasNext(){return iter.hasNext()||count>0;}
                public T next()
                {
                    if(count==0)
                    {
                        curElement=iter.next();
                        count=get(curElement);
                    }
                    count--; return curElement;
                }
            };
        }
    }
    static long abs(long x){
        if(x<0) x*=-1;
        return x;
    }

    public static void main(String[] args) {
        MyScanner sc = new MyScanner();
        int t = 1;
        while (t-- != 0) {
            int n = sc.nextInt();
            int a[] = new int [n];
            for(int i=0;i<n;i++)
                a[i] = sc.nextInt();
            long dp[][] = new long[n][n];
            int prev = -1;
            for(int i =0;i<n;i++){
                if(a[i] == 0) continue;
                long prevSum = Long.MAX_VALUE;
                for(int j=0;j<n;j++){
                    if(a[j] == 0){
                        dp[i][j] =  abs(i - j);
                        if(prev != -1){ // not the first 1
                            if(prevSum != Long.MAX_VALUE)dp[i][j] += prevSum;
                            else dp[i][j] = prevSum;
                            prevSum = Math.min(prevSum, dp[prev][j]);
                        }
                    }
                }
                prev = i;
            }
            long ans = Long.MAX_VALUE;
            if(prev != -1)
                for(int i =0;i<n;i++){
                    if(a[i] == 0)
                     ans = Math.min(ans,dp[prev][i]);
                }
            else
                ans = 0;
            System.out.println(ans);


        }
    }
}
