Problem: You are given an array a consisting of n positive integers. You pick two integer numbers l and r from 1 to n, inclusive (numbers are picked randomly, equiprobably and independently). If l > r, then you swap values of l and r. You have to calculate the expected value of the number of unique elements in segment of the array from index l to index r, inclusive (1-indexed).

Input

The first line contains one integer number n (1 ≤ n ≤ 106). The second line contains n integer numbers a1, a2, ... an (1 ≤ ai ≤ 106) — elements of the array.

Output

Print one number — the expected number of unique elements in chosen segment. 

Your answer will be considered correct if its absolute or relative error doesn't exceed 10 - 4 — formally, the answer is correct if <image>, where x is jury's answer, and y is your answer.

Examples

Input

2
1 2


Output

1.500000


Input

2
2 2


Output

1.000000

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.io.PrintWriter;
import java.util.ArrayList;
public class F846 
{
    static class Scanner
    {
        BufferedReader br;
        StringTokenizer tk=new StringTokenizer("");
        public Scanner(InputStream is) 
        {
            br=new BufferedReader(new InputStreamReader(is));
        }
        public int nextInt() throws IOException
        {
            if(tk.hasMoreTokens())
                return Integer.parseInt(tk.nextToken());
            tk=new StringTokenizer(br.readLine());
            return nextInt();
        }
        public long nextLong() throws IOException
        {
            if(tk.hasMoreTokens())
                return Long.parseLong(tk.nextToken());
            tk=new StringTokenizer(br.readLine());
            return nextLong();
        }
        public String next() throws IOException
        {
            if(tk.hasMoreTokens())
                return (tk.nextToken());
            tk=new StringTokenizer(br.readLine());
            return next();
        }
        public String nextLine() throws IOException
        {
            tk=new StringTokenizer("");
            return br.readLine();
        }
        public double nextDouble() throws IOException
        {
            if(tk.hasMoreTokens())
                return Double.parseDouble(tk.nextToken());
            tk=new StringTokenizer(br.readLine());
            return nextDouble();
        }
        public char nextChar() throws IOException
        {
            if(tk.hasMoreTokens())
                return (tk.nextToken().charAt(0));
            tk=new StringTokenizer(br.readLine());
            return nextChar();
        }
        public int[] nextIntArray(int n) throws IOException
        {
            int a[]=new int[n];
            for(int i=0;i<n;i++)
                a[i]=nextInt();
            return a;
        }
        public long[] nextLongArray(int n) throws IOException
        {
            long a[]=new long[n];
            for(int i=0;i<n;i++)
                a[i]=nextLong();
            return a;
        }
        public int[] nextIntArrayOneBased(int n) throws IOException
        {
            int a[]=new int[n+1];
            for(int i=1;i<=n;i++)
                a[i]=nextInt();
            return a;
        }
        public long[] nextLongArrayOneBased(int n) throws IOException
        {
            long a[]=new long[n+1];
            for(int i=1;i<=n;i++)
                a[i]=nextLong();
            return a;
        }
    
    
    }
    
    public static void main(String args[]) throws IOException
    {
        Scanner in=new Scanner(System.in);
        PrintWriter out=new PrintWriter(System.out);
        int n=in.nextInt();
        ArrayList<Integer> a[]=new ArrayList[1000001];
        for(int i=1;i<=1000000;i++)
            a[i]=new ArrayList<>();
        for(int i=1;i<=1000000;i++)
            a[i].add(0);
        for(int i=1;i<=n;i++)
            a[in.nextInt()].add(i);
        for(int i=1;i<=1000000;i++)
            a[i].add(n+1);
        long ans=0;
        ans+=n;
        for(long i=2;i<=n;i++)
            ans+=2*(i*(n-i+1));
        for(int i=1;i<=1000000;i++)
        {
            if(a[i].size()>3)
            {
                long sum=(n+1)*(long)(a[i].size()-3);
                for(int j=2;j<a[i].size()-1;j++)
                    sum-=a[i].get(j);
                for(int j=1;j<a[i].size()-1;j++)
                {
                    ans-=2*sum*(a[i].get(j)-a[i].get(j-1));
                    sum+=a[i].get(j+1);
                    sum-=(n+1);
                }
                
            }
        }
        double v=ans;
        v/=n;
        v/=n;
        out.println(v);
        out.close();
    }

}