Problem: Andrew and Jerry are playing a game with Harry as the scorekeeper. The game consists of three rounds. In each round, Andrew and Jerry draw randomly without replacement from a jar containing n balls, each labeled with a distinct positive integer. Without looking, they hand their balls to Harry, who awards the point to the player with the larger number and returns the balls to the jar. The winner of the game is the one who wins at least two of the three rounds.

Andrew wins rounds 1 and 2 while Jerry wins round 3, so Andrew wins the game. However, Jerry is unhappy with this system, claiming that he will often lose the match despite having the higher overall total. What is the probability that the sum of the three balls Jerry drew is strictly higher than the sum of the three balls Andrew drew?

Input

The first line of input contains a single integer n (2 ≤ n ≤ 2000) — the number of balls in the jar.

The second line contains n integers ai (1 ≤ ai ≤ 5000) — the number written on the ith ball. It is guaranteed that no two balls have the same number.

Output

Print a single real value — the probability that Jerry has a higher total, given that Andrew wins the first two rounds and Jerry wins the third. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6. 

Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.

Examples

Input

2
1 2


Output

0.0000000000


Input

3
1 2 10


Output

0.0740740741

Note

In the first case, there are only two balls. In the first two rounds, Andrew must have drawn the 2 and Jerry must have drawn the 1, and vice versa in the final round. Thus, Andrew's sum is 5 and Jerry's sum is 4, so Jerry never has a higher total.

In the second case, each game could've had three outcomes — 10 - 2, 10 - 1, or 2 - 1. Jerry has a higher total if and only if Andrew won 2 - 1 in both of the first two rounds, and Jerry drew the 10 in the last round. This has probability <image>.

Solution (in JAVA):

Solution:

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;

public class D {
	
	// maximum number that an element in an array can receive
	static int MAX_NUM = 5005;
	
	public static void main(String [] args) throws IOException {		
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		
		int N = Integer.parseInt(br.readLine());
		StringTokenizer tokenizer = new StringTokenizer(br.readLine());
		
		int [] a = new int[N];
		for(int i=0; i<N; i++) { 
			a[i] = Integer.parseInt(tokenizer.nextToken());
		}
		// sort the array to find the differences between the winner and the loser
		// more easily
		Arrays.sort(a);
		
		
		// diff[i] - how many times difference of 'i' units occurs when the points of the loser
		// are substracted from the points of the winner
		long [] diff = new long[MAX_NUM];
		
		for(int i=0; i<N; i++) 
			for(int j=i+1; j<N; j++)
				diff[a[j] - a[i]]++;
		
		
		// this array will represent the sum of differences in the first two rounds
		// sumDiff[i] - how many ways to get to sum of 'i' units (advantage of Andrew over Jerry)
		
		// sumDiff[i] - number of outcomes such that after two rounds Andrew leads by 'i' points
		long [] sumDiff = new long[2*MAX_NUM];
		
		for(int i=0; i<MAX_NUM; i++) 
			for(int j=0; j<MAX_NUM; j++) 
				sumDiff[i + j] += diff[i] * diff[j];
		
		
		long total = 0; 
		for(int i=0; i<MAX_NUM-1; i++) {			
			if(sumDiff[i] == 0) continue;
			
			
			// if Andrew leads by 'i' points, let's count how many ways are there
			// such that Jerry wins with a difference by more than 'i' points
			int howManyGreater = 0;
			
			for(int j=1+i; j<MAX_NUM; j++) { 
				howManyGreater += diff[j];
			}
			
			
			// number of ways Jerry can score a difference greater than i in the third round
			// * 
			// number of ways Andrew can score a total difference of 'i', after two rounds
			total += howManyGreater * sumDiff[i];
		}
		
		double nom = total;
		double denom = 1;
		
		
		// There are (N*(N-1) / 2)^3 possible outcomes
		// because in each of the three (^3) rounds we have N*(N-1) possible outcomes
		// and we assume that the drawings in each round are independent
		
		for(int i=0; i<3; i++) { 
			denom *= (N * (N-1))/2.0;
		}
		
		System.out.println(nom/denom);
	}

}