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.
/*
    JAI MATA DI
 */

import java.util.*;

import javax.print.attribute.HashAttributeSet;


import java.io.*;
import java.math.BigInteger;
public class CP {
	  static class FR{
			BufferedReader br;
			StringTokenizer st;
			public FR() {
				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 int mod  = 1000000007;
	 static int gcd(int a, int b)
	  {      
	     if (b == 0)
	        return a;
	     return gcd(b, a % b); 
	  }
	 static long gcd(long a, long b)
	  {      
	     if (b == 0)
	        return a;
	     return gcd(b, a % b); 
	  }
	 static boolean[] prime(int num) {
			boolean[] bool = new boolean[num];
		     
		      for (int i = 0; i< bool.length; i++) {
		         bool[i] = true;
		      }
		      for (int i = 2; i< Math.sqrt(num); i++) {
		         if(bool[i] == true) {
		            for(int j = (i*i); j<num; j = j+i) {
		               bool[j] = false;
		            }
		         }
		      }
		      if(num >= 0) {
		    	  bool[0] = false;
		    	  bool[1] = false;
		      }
		      
		      return bool;
		}
	 static class Pair implements Comparable<Pair>{
	 		int ind;
	 		int val;
	 		Pair(int key , int value){
	 			this.ind = key;
	 			this.val = value;
	 		}
			@Override
			public int compareTo(Pair o) {
				return this.val - o.val;
			}
//			@Override
//			public int hashCode(){
//				return first + second;
//			}
	 	}
	 	
	 
	 

/* ***************************************************************************************************************************************************/	 
	 static FR sc = new FR();
	 static StringBuilder sb = new StringBuilder();
	 public static void main(String args[]) {
//		 int tc = sc.nextInt();
//		 while(tc-- > 0) {
			 TEST_CASE();
//		 }
			System.out.println(sb);
		 }		

		static void TEST_CASE() {
			int n = sc.nextInt();
			int[] arr = new int[n];	
			for(int i = 0 ; i<n ;i++) arr[i] = sc.nextInt();
			int[] ch = new int[n];
			long ans = 0;
			for(int i = 0 ; i < n ; i++) {
				if(arr[i] == 0) continue;
				int l = i-1 , r = i+1;
				while(l>=0) {
					if(arr[l] == 0 && ch[l] == 0) {
						break;
					}
					l--;
				}
				while(r<n) {
					if(arr[r] == 0 && ch[r] == 0) {
						break;
					}
					r++;
				}
				if(l <0) {
					ch[r] = 1;
					ans += Math.abs(r-i);
				}else if(r>=n) {
					ch[l] = 1;
					ans += Math.abs(l-i);
				}else {
					if(Math.abs(l-i)<=Math.abs(r-i)) {
						ch[l] = 1;
						ans += Math.abs(l-i);
					}else {
						ch[r] = 1;
						ans += Math.abs(r-i);
					}
				}
			}
//			6
//			0 0 1 1 0 1
			long ans2 = 0;
			LinkedList<Integer> l1 = new LinkedList<Integer>();
			LinkedList<Integer> l2 = new LinkedList<Integer>();
			LinkedList<Integer> l3 = new LinkedList<Integer>();
			for(int i = 0 ; i<n ; i++) {
				if(arr[i] == 1) l1.add(i);
				if(ch[i] == 1) {
					l2.add(i);
					l3.addLast(i);
				}
				
			}
			long ans3 = 0;
			while(!l1.isEmpty()) {
				int a = l1.removeFirst();
				ans2 += (long)Math.abs(a - l2.removeFirst());
				ans3 += (long)Math.abs(a - l3.removeFirst());
			}
			
			ans2 = Math.min(ans2, ans3);
			sb.append(Math.min(ans, ans2)+"\n");
			
			
		}
		
	
	

}















