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.
//https://github.com/EgorKulikov/yaal/tree/master/lib/main/net/egork
        
import java.util.*;
import java.io.*;
public class A{
    static PrintWriter out;
    static InputReader in;
    public static void main(String args[]){
        out = new PrintWriter(System.out);
        in = new InputReader();
        new A();
        out.flush(); out.close();
    }   
    A(){
        solve();
    }
    final int maxn = 5001, maxv = maxn * maxn + 100;
    void solve(){
        // int t = in.nextInt();
        int t = 1;
        while(t-- > 0){
            int n = in.nextInt();
            int a[][] = new int[n + 1][2], cnt[] = new int[2];
            for(int i = 1; i <= n; i++){
                int x = in.nextInt();
                cnt[x]++;
                a[cnt[x]][x] = i;
            }
            int dp[][] = new int[cnt[1] + 1][cnt[0] + 1];
            for(int i = 0; i <= cnt[1]; i++){
                for(int j = 0; j <= cnt[0]; j++){
                    dp[i][j] = maxv;
                }
            }
            dp[0][0] = 0;
            for(int i = 0; i <= cnt[1]; i++){
                for(int j = i; j <= cnt[0]; j++){
                    if(j == i && j == 0)continue;
                    dp[i][j] = Math.min(dp[i][j - 1], i > 0 ? dp[i - 1][j - 1] + Math.abs(a[i][1] - a[j][0]) : maxv);
                }
            }
            out.print(dp[cnt[1]][cnt[0]]);
        }
    }
    public static class InputReader{
        BufferedReader br;
        StringTokenizer st;
        InputReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        public int nextInt(){
            return Integer.parseInt(next());
        }
        public long nextLong(){
            return Long.parseLong(next());
        }
        public double nextDouble(){
            return Double.parseDouble(next());
        }
        public String next(){
            while(st == null || !st.hasMoreTokens()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch(IOException e){}
            }
            return st.nextToken();
        }
    }
}
	