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.awt.*;
import java.io.*;
import java.sql.Array;
import java.util.*;
import java.util.List;

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

        public FastReader()
        {
            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 final int N=200005;
    static final int mod=1000000007;
    static final int INF=1000000009;
    static final long oo=1000000000000000009l;
    static final int numBit=17;
    static FastReader r=new FastReader();
//    static PrintWriter pw = new PrintWriter(System.out);
    static class SegmentTree{
        int []seg;
        int []lazy;
        SegmentTree(int n){
            seg=new int[n*4];
            lazy=new int[n*4];
//            Arrays.fill(seg,0);
//            Arrays.fill(lazy,0);
        }
        public int min(int a,int b){
            return Math.min(a,b);
        }
        public void pushdown(int id){
            long val=lazy[id];
            seg[id*2]+=val;
            lazy[id*2]+=val;
            seg[id*2+1]+=val;
            lazy[id*2+1]+=val;
            lazy[id]=0;
        }
        public void update(int id,int l,int r,int beg,int fin,long val){
            if(fin<l||r<beg) return;
            if(beg<=l&&r<=fin){
                seg[id]+=val;
                lazy[id]+=val;
                return;
            }
            int mid=(l+r)>>1;
            pushdown(id);
            update(id*2,l,mid,beg,fin,val);
            update(id*2+1,mid+1,r,beg,fin,val);
            seg[id]=seg[id*2]+seg[id*2+1];
        }
        public int getSum(int id,int l,int r,int beg,int fin){
            if(fin<l||r<beg) return 0;
            if(beg<=l&&r<=fin){
                return seg[id];
            }
            int mid=(l+r)>>1;
            pushdown(id);
            return getSum(id*2,l,mid,beg,fin)+getSum(id*2+1,mid+1,r,beg,fin);
        }
    }

    static class Fenwick{
        long[] f;
        int n;
        Fenwick(int n){
            this.n=n;
            f=new long[n];
        }
        public void add(int pos,long val){
            for(;pos<n;pos+=pos&-pos){
                f[pos]=Math.max(f[pos],val);
            }
        }

        public long query(int pos){
            long ret=0;
            for(;pos>0;pos-=pos&-pos){
                ret=Math.max(ret,f[pos]);
            }
            return ret;
        }
    }

    public static void main(String[] args) throws IOException {
        int n=r.nextInt();
        int[] a=new int[n];
        for(int i=0;i<n;++i) a[i]=r.nextInt();
        List<Integer> occ=new ArrayList<>(),free=new ArrayList<>();
        occ.add(0);
        free.add(0);
        for(int i=0;i<n;++i){
            if(a[i]==1) occ.add(i+1);
            else free.add(i+1);
        }
        int m=occ.size()-1;
        n=free.size()-1;
        int[][] dp=new int[n+1][m+1];
        for(int i=0;i<=n;++i){
            for(int j=0;j<=m;++j){
                if(j>i) dp[i][j]=INF;
            }
        }
//        System.out.println(dp[0][1]);
        for(int i=1;i<=n;++i){
            for(int j=1;j<=m;++j){
                if(j>i) dp[i][j]=INF;
                else dp[i][j]=Math.min(dp[i-1][j],dp[i-1][j-1]+Math.abs(occ.get(j)-free.get(i)));
//                if(i==1&&j==1){
//                    System.out.println("dp i j-1 : "+dp[i-1][j]);
//                    System.out.println("dp 1 1 = "+dp[i][j]);
//                }
            }
        }
        System.out.println(dp[n][m]);
//        pw.close();
    }
}
