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;
struct Fast {Fast(){std::cin.tie(0);ios::sync_with_stdio(false);cout.precision(20);}} fast;

/* define */
#define FOR(I,X,Y) for(long long (I)=(X);(I)<(Y);(I)++)
#define REP(I,X,Y) for(long long (I)=(Y)-1;(I)>=(X);(I)--)
#define ALL(X) (X).begin(),(X).end()
#define pb push_back
#define COUNT(V,X) (upper_bound((V).begin(),(V).end(),X)-lower_bound((V).begin(),(V).end(),X))
#define debug(x) cerr<<#x<<':'<<x<<endl;
#define DEBUG(v) cerr<<#v<<':';for(auto xXx:v)cerr<<xXx<<' ';cerr<<endl;
#define Yes(X) if(X){cout<<"Yes"<<endl;}else{cout<<"No"<<endl;}
#define YES(X) if(X){cout<<"YES"<<endl;}else{cout<<"NO"<<endl;}
#define ctoi(C) (C-'0')

/* alias */
using ll = long long;
using ld = long double;
using vi = vector<int>;
using vii = vector<vector<int>>;
using vl = vector<long long>;
using vll = vector<vector<long long>>;
using pi = pair<int,int>;
using pl = pair<long long,long long>;
template<typename T> using PQ = priority_queue<T>;
template<typename T> using minPQ = priority_queue<T, vector<T>, greater<T>>;

/* const */
const long long dx[] = {1,0,-1,0};
const long long dy[] = {0,1,0,-1};
const long long dx8[] = {1,1,0,-1,-1,-1,0,1};
const long long dy8[] = {0,1,1,1,0,-1,-1,-1};
const long long dx9[] = {1,1,0,-1,-1,-1,0,1,0};
const long long dy9[] = {0,1,1,1,0,-1,-1,-1,0};
const int INF = 1000000007;
const long long LINF = 1000000000000000007;

/* func */
template <typename T> inline bool chmin(T& a, const T& b) {if (a > b) a = b; return a > b;}
template <typename T> inline bool chmax(T& a, const T& b) {if (a < b) a = b; return a < b;}
long long max(long long x,int y){return max(x,(long long)y);}
long long max(int x,long long y){return max((long long)x,y);}
long long min(long long x,int y){return min(x,(long long)y);}
long long min(int x,long long y){return min((long long)x,y);}

/* liblary */



/* main */

ld memo[500][500][500];
ld rec(ll i,ll j,ll k,const ll& N){
	if(memo[i][j][k])return memo[i][j][k];
	if(!i&&!j&&!k)return 0;
	if(i == 1 && !j && !k)return (ld)N/i;
	ld ans = 0;
	if(i)ans += i*rec(i-1,j,k,N);
	if(j)ans += j*rec(i+1,j-1,k,N);
	if(k)ans += k*rec(i,j+1,k-1,N);
	return memo[i][j][k] = (ld)(ans+N)/(i+j+k);
}

/*
memo[i][j][k] =
((N-i-j-k)/N)*memo[i][j][k]+
(i/N)*memo[i-1][j][k]+
(j/N)*memo[i+1][j-1][k]+
(k/N)*memo[i][j+1][k-1]+
1
*/

signed main(){
	ll N;
	cin >> N;
	vl a(N);
	ll counter[3] = {0};
	FOR(i,0,N){
		ll tmp;
		cin >> tmp;
		if(tmp)counter[tmp-1]++;
	}
	cout << rec(counter[0],counter[1],counter[2],N) << endl;
}