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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

/*
author: madhav_1999 aka orthodoxparadox
18-Jan-2019
00:33:40
*/

#include <bits/stdc++.h>
#include <ext/pb_ds/detail/standard_policies.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

#define dbl long double
#define int ll
#define ll long long
#define pii pair<int, int>
#define fup(i, begin, end) for (__typeof(end) i = (begin) - ((begin) > (end)); i != (end) - ((begin) > (end)); i += 1 - 2 * ((begin) > (end)))
#define fnu for(ll i = 0; i < n; i++)
#define rev(Y) reverse(all(Y))
#define oiint ostream_iterator<int>(cout, " ") 
#define pb emplace_back
#define mp make_pair
#define ff first
#define ss second
#define endl '\n'
#define all(a) a.begin(), a.end()
#define initialise(a, x) memset(a, x, sizeof(a))
#define onlyunique(v) v.erase(unique(all(v)), v.end()); //only for sorted vector
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize ("O3")
#pragma GCC optimize ("O2")
#pragma GCC optimize("Ofast")
#pragma GCC optimize ("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#define _CRT_SECURE_NO_WARNINGS
#ifdef __APPLE__
#define dbg(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){cerr << name << " : " << arg1 << endl;}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
	const char* comma = strchr(names + 1, ',');
	cerr.write(names, comma - names) << " : " << arg1 << "  ";
	__f(comma + 1, args...);}
#else
#define dbg(...)
#endif
template <typename X> ostream& operator << (ostream& x,const vector<X>& v){for(int i=0;i<v.size();++i) x<<v[i]<<" ";return x;} 
template <typename X> ostream& operator << (ostream& x,const set<X>& v){for(auto it:v) x<<it<<" ";return x;} 
template <typename X, typename Y> ostream& operator << (ostream& x, const pair<X,Y>& v) {x<<v.ff<<" "<<v.ss;return x;} 
typedef tree <int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
struct pair_hash {inline std::size_t operator()(const std::pair<int, int> & v) const {return v.first*31+v.second;}};
const int mod = 1e9 + 7;

int takemod(int a){return ((a%mod)+mod)%mod;}

int fast_exp(int base, int expo) {
	int res=1;
	while(expo>0) {
		if(expo&1) res=(res*base)%mod;
		base=(base*base)%mod;
		expo>>=1;}
	return res;}

int modinv(int a){return takemod(fast_exp(takemod(a), mod-2));}

const int N = 301;
dbl dp[N][N][N];
dbl eps = 1e-12;
int n;

dbl solve(int c1, int c2, int c3)
{
	if(c1 == 0 and c2 == 0 and c3 == 0)
	{
		return 0;
	}
	if(dp[c1][c2][c3] != 0)
	{
		return dp[c1][c2][c3];
	}
	dbl p1 = (dbl)c1 / n;
	dbl p2 = (dbl)c2 / n;
	dbl p3 = (dbl)c3 / n;
	dbl px1 = (dbl)c1 / (c1 + c2 + c3);
	dbl px2 = (dbl)c2 / (c1 + c2 + c3);
	dbl px3 = (dbl)c3 / (c1 + c2 + c3);
	dbl r1 = 0, r2 = 0, r3 = 0;
	if (c1)
	{
		r1 = px1 * (solve(c1 - 1, c2, c3));
	}
	if (c2)
	{
		r2 = px2 * (solve(c1 + 1, c2 - 1, c3));
	}
	if (c3)
	{
		r3 = px3 * (solve(c1, c2 + 1, c3 - 1));
	}
	dbl expe = (dbl)(c1 + c2 + c3) / n;
	expe = 1/expe;
	return dp[c1][c2][c3] = r1 + r2 + r3 + expe;
}

signed main(){
	ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
	#ifdef __APPLE__
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
	freopen("error.txt", "w", stderr);
	#endif
	clock_t t1, t2;
	t1 = clock();


	cin >> n;
	int cnt[4]{};
	for(int i = 0; i < n; i++)
	{
		int x;
		cin >> x;
		cnt[x]++;
	}
	cout << fixed << setprecision(20);
	cout << solve(cnt[1], cnt[2], cnt[3]);
	t2 = clock(); cerr << "time taken: " << (t2-t1)/(dbl)CLOCKS_PER_SEC;
	return 0;
}