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)

#include<bits/stdc++.h>
using namespace std;

#define ll             long long
#define LL             long long
#define pii            pair<int, int>
#define pll            pair<ll, ll>
#define pil            pair<int, ll>
#define pli            pair<ll, int>
#define vi             vector<int>
#define vl             vector<ll>
#define vii            vector<pii>
#define vll            vector<pll>
#define vil            vector<pil>
#define vli            vector<pli>
#define pb             push_back
#define ppb            pop_back
#define mp             make_pair
#define ff             first
#define ss             second
#define sz(x)          (int)x.size()
#define TN             typename
#define all(v)         v.begin(), v.end()
#define fill(a, b)     memset(a, b, sizeof(a))
#define endl           '\n'

template <TN T> T gcd(T a, T b) {return !b ? a : gcd(b, a % b);}
template <TN T> T lcm(T a, T b) {return a * (b / gcd(a, b));}
template <TN T> T sqr(T a) {return a * a;}
template <TN T> T cube(T a) {return a * a * a;}
template <TN T> inline void smin(T &a, T b) {a = a < b ? a : b;}
template <TN T> inline void smax(T &a, T b) {a = a > b ? a : b;}

template <TN T> inline void Int(T &n) {
	n = 0; int f = 1; register int ch = getchar();
	for (; !isdigit(ch); ch = getchar()) if (ch == '-') f = -1;
	for (; isdigit(ch); ch = getchar()) n = (n << 3) + (n << 1) + ch - '0';
	n = n * f;
}

#define error(args...) {vector <string> _v = split(#args, ','); err(_v.begin(), args); cout << endl;}
vector <string> split(const string &s, char c) {
	vector <string> v;
	stringstream ss(s); string x;
	while (getline(ss, x, c)) v.emplace_back(x);
	return move(v);
}
void err(vector <string> :: iterator it) {}
template <typename T, typename... Args>
void err(vector <string> :: iterator it, T a, Args... args) {
	cout << it -> substr((*it)[0] == ' ', it -> length()) << " = " << a << ", ";
	err(++it, args...);
}

inline int in() {int n; scanf("%d", &n); return n;}
inline ll Lin() {ll n; scanf("%lld", &n); return n;}
inline double Din() {double n; scanf("%lf", &n); return n;}

inline int add(int a, int b, int mod) {a += b; return a >= mod ? a - mod : a;}
inline int sub(int a, int b, int mod) {a -= b; return a < 0 ? a + mod : a;}
inline int mul(int a, int b, int mod) {return (ll)a * b % mod;}

const int mod        = (int)1e9 + 7;
const ll  Mod        = (ll)1e9 + 7;

inline int add(int a, int b) {a += b; return a >= mod ? a - mod : a;}
inline int sub(int a, int b) {a -= b; return a < 0 ? a + mod : a;}
inline int mul(int a, int b) {return (ll)a * b % mod;}

const int inf        = (int)2e9 + 5;
const ll  Inf        = (ll)1e18 + 5;
const int N          = 3e2 + 5;


double p[N][N][N], e[N][N][N];
int cnt[4];


int solve() {
	int n = in();
	for (int i = 0; i < n; i++) {
		int x = in();
		cnt[x]++;
	}
	p[cnt[1]][cnt[2]][cnt[3]] = 1.0;
	for (int c = n; c >= 0; c--) {
		for (int b = n; b >= 0; b--) {
			for (int a = n; a >= 0; a--) {
				if (a + b + c == 0 || a + b + c > n) continue;
				double waste = (double) (n - (a + b + c)) / n;
				double e_move = waste / (1.0 - waste) + 1.0;
				e[a][b][c] += e_move * p[a][b][c];
				if (a > 0) {
					double go = (double) a / (a + b + c);
					p[a - 1][b][c] += p[a][b][c] * go;
					e[a - 1][b][c] += e[a][b][c] * go;
				}

				if (b > 0) {
					double go = (double) b / (a + b + c);
					p[a + 1][b - 1][c] += p[a][b][c] * go;
					e[a + 1][b - 1][c] += e[a][b][c] * go;
				}

				if (c > 0) {
					double go = (double) c / (a + b + c);
					p[a][b + 1][c - 1] += p[a][b][c] * go;
					e[a][b + 1][c - 1] += e[a][b][c] * go;
				}
			}
		}
	}
	printf("%.10f\n", e[0][0][0]);
	return 0;
}

int main() {
	int test = 1, tc = 0;
	//scanf("%d", &test);
	while (test--) {
		//printf("Case %d: ", ++tc);
		solve();
	}
	return 0;
}