There are N dishes, numbered 1, 2, \ldots, N. Initially, for each i (1 \leq i \leq N), Dish i has a_i (1 \leq a_i \leq 3) pieces of sushi on it.

Taro will perform the following operation repeatedly until all the pieces of sushi are eaten:

* Roll a die that shows the numbers 1, 2, \ldots, N with equal probabilities, and let i be the outcome. If there are some pieces of sushi on Dish i, eat one of them; if there is none, do nothing.



Find the expected number of times the operation is performed before all the pieces of sushi are eaten.

Constraints

* All values in input are integers.
* 1 \leq N \leq 300
* 1 \leq a_i \leq 3

Input

Input is given from Standard Input in the following format:


N
a_1 a_2 \ldots a_N


Output

Print the expected number of times the operation is performed before all the pieces of sushi are eaten. The output is considered correct when the relative difference is not greater than 10^{-9}.

Examples

Input

3
1 1 1


Output

5.5


Input

1
3


Output

3


Input

2
1 2


Output

4.5


Input

10
1 3 2 3 3 2 3 2 1 3


Output

54.48064457488221
import java.util.Arrays;
import java.util.Scanner;

public class Main {
    static final Scanner sc    = new Scanner(System.in);
    static final int     MOD   = (int) 1E9 + 7;
    
    public static void main(String[] args) {
        int[] a = new int[4];
        int N = nint();
        
        for (int i = 0; i < N; i++) {
            a[nint()]++;
        }
        
        double[][][] dp = new double[N+2][N+2][N+2];
        
        for (int i3 = 0; i3 <= N; i3++) {
            for (int i2 = 0; i2 <= N; i2++) {
                for (int i1 = 0; i1 <= N; i1++) {
                    int s = i1 + i2 + i3;
                    if (s == 0) continue;
                    
                    double temp = N;
                    if (i1 > 0) temp += i1 * dp[i1-1][i2  ][i3  ];
                    if (i2 > 0) temp += i2 * dp[i1+1][i2-1][i3  ];
                    if (i3 > 0) temp += i3 * dp[i1  ][i2+1][i3-1];
                    dp[i1][i2][i3] = temp / s;
                }
            }
        }
        
        System.out.println(dp[a[1]][a[2]][a[3]]);
    }
    
    @Deprecated
    static int nint() {
        return sc.nextInt();
    }

    @Deprecated
    static int[] nints(int N) {
        return nints(N, 0, 0);
    }

    @Deprecated
    private static int[] nints(int N, int padL, int padR) {
        int[] a = new int[padL + N + padR];
        for (int i = 0; i < N; i++)
            a[padL + i] = nint();
        return a;
    }

    static long nlong() {
        return sc.nextLong();
    }

    static long[] nlongs(int N) {
        return nlongs(N, 0, 0);
    }

    static long[] nlongs(int N, int padL, int padR) {
        long[] a = new long[padL + N + padR];
        for (int i = 0; i < N; i++)
            a[padL + i] = nlong();
        return a;
    }

    static double ndouble() {
        return sc.nextDouble();
    }

    static double[] ndoubles(int N) {
        return ndoubles(N, 0, 0);
    }

    static double[] ndoubles(int N, int padL, int padR) {
        double[] d = new double[N + padL + padR];
        for (int i = 0; i < N; i++) {
            d[padL + i] = ndouble();
        }
        return d;
    }

    static String nstr() {
        return sc.next();
    }

    static char[] nchars() {
        return sc.next().toCharArray();
    }

    static char[] nchars(int padL, int padR) {
        char[] temp = sc.next().toCharArray();
        char[] ret = new char[temp.length + padL + padR];
        System.arraycopy(temp, 0, ret, padL, temp.length);
        return ret;
    }

    static char[][] nchars2(int H, int W) {
        return nchars2(H, W, 0, 0);
    }

    static char[][] nchars2(int H, int W, int padLU, int padRD) {
        char[][] a2 = new char[H + padLU + padRD][W + padLU + padRD];
        for (int i = 0; i < H; i++)
            System.arraycopy(nchars(), 0, a2[padLU + i], padLU, W);
        return a2;
    }

    static long min(long... ls) {
        return Arrays.stream(ls).min().getAsLong();
    }

    static long max(long... ls) {
        return Arrays.stream(ls).max().getAsLong();
    }

    static long abs(long a) {
        return Math.abs(a);
    }
}
