Problem:
Score : 100 points

Problem Statement

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}.

Sample Input 1

3
1 1 1

Sample Output 1

5.5

The expected number of operations before the first piece of sushi is eaten, is 1.
After that, the expected number of operations before the second sushi is eaten, is 1.5.
After that, the expected number of operations before the third sushi is eaten, is 3.
Thus, the expected total number of operations is 1 + 1.5 + 3 = 5.5.

Sample Input 2

1
3

Sample Output 2

3

Outputs such as 3.00, 3.000000003 and 2.999999997 will also be accepted.

Sample Input 3

2
1 2

Sample Output 3

4.5

Sample Input 4

10
1 3 2 3 3 2 3 2 1 3

Sample Output 4

54.48064457488221

Solution in Python (3.4.3) (judge verdict: Time Limit Exceeded)

import sys
import math
from decimal import Decimal, ROUND_HALF_UP, ROUND_HALF_EVEN
from collections import deque
from bisect import bisect_left
from itertools import product
def I(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LI2(N): return [list(map(int, sys.stdin.readline().split())) for i in range(N)]
#文字列を一文字ずつ数字に変換、'5678'を[5,6,7,8]とできる
def LSI(): return list(map(int, list(sys.stdin.readline().rstrip())))
def LSI2(N): return [list(map(int, list(sys.stdin.readline().rstrip()))) for i in range(N)]
#文字列として取得
def ST(): return sys.stdin.readline().rstrip()
def LST(): return sys.stdin.readline().rstrip().split()
def LST2(N): return [sys.stdin.readline().rstrip().split() for i in range(N)]
def FILL(i,h): return [i for j in range(h)]
def FILL2(i,h,w): return [FILL(i,w) for j in range(h)]
def FILL3(i,h,w,d): return [FILL2(i,w,d) for j in range(h)]
def FILL4(i,h,w,d,d2): return [FILL3(i,w,d,d2) for j in range(h)]
def sisha(num,digit): return Decimal(str(num)).quantize(Decimal(digit),rounding=ROUND_HALF_UP)
#'0.01'や'1E1'などで指定、整数に戻すならintをかます
MOD = 1000000007
INF = float("inf")
sys.setrecursionlimit(10**6+10)

N = I()
A = LI()
one = 0
two = 0
three = 0
for a in A:
    if a==1:
        one += 1
    elif a==2:
        two += 1
    elif a==3:
        three += 1

dp = FILL3(-1.0,N+1,N+1,N+1)
dp[0][0][0] = 0
def rec(i,j,k):
    if i<0 or j<0 or k<0:
        return 0
    if dp[i][j][k] != -1.0:
        return dp[i][j][k]
    zero = N-i-j-k
    p_0 = zero/N
    p_1 = i/N
    p_2 = j/N
    p_3 = k/N
    ret = p_1*rec(i-1,j,k) + p_2*rec(i+1,j-1,k) + p_3*rec(i,j+1,k-1) + 1
    ret /= 1-p_0
    dp[i][j][k] = ret
    return ret

print(rec(one,two,three))