#include <string>
#include <vector>
#include <algorithm>
#include <numeric>
#include <set>
#include <map>
#include <queue>
#include <iostream>
#include <sstream>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <cctype>
#include <cassert>
#include <limits>
#include <functional>
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i))
#define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i))
#if defined(_MSC_VER) || __cplusplus > 199711L
#define aut(r,v) auto r = (v)
#else
#define aut(r,v) __typeof(v) r = (v)
#endif
#define each(it,o) for(aut(it, (o).begin()); it != (o).end(); ++ it)
#define all(o) (o).begin(), (o).end()
#define pb(x) push_back(x)
#define mp(x,y) make_pair((x),(y))
#define mset(m,v) memset(m,v,sizeof(m))
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3fLL
using namespace std;
typedef vector<int> vi; typedef pair<int,int> pii; typedef vector<pair<int,int> > vpii; typedef long long ll;
template<typename T, typename U> inline void amin(T &x, U y) { if(y < x) x = y; }
template<typename T, typename U> inline void amax(T &x, U y) { if(x < y) x = y; }

//phi(P) | Q
//r^Q = 1

//p_i - 1 = 2^{e_i} × q_i
//p_i | h
//p_i | (c - 1)
//c ≡ 1 (mod p_i) 
//max_j {f_j(r)} = f_i(r)

//  { a^{2^k} = 1 | a <- S }
//= { 2^k×l ≡ 0 (mod p-1) | l <- [0..p-2] }
//= { v_2(l) >= v_2(p-1) - k | l <- [0..p-2] }
//= (p-1) / 2^{max(0, v_2(p-1) - k)}

//

int main() {
	int T;
	scanf("%d", &T);
	rep(ii, T) {
		int N;
		scanf("%d", &N);
		vector<long long> P(N);
		rep(i, N)
			scanf("%lld", &P[i]);
		vector<int> v_2(N);
		rep(i, N) {
			long long x = P[i] - 1;
			int k = 0;
			while(x % 2 == 0) x /= 2, ++ k;
			v_2[i] = k;
		}
		int maxv2 = *max_element(all(v_2));
		vector<double> dp(1 << N);
		dp[0] = 1;
		rep(i, N) dp[1 << i] = 1;
		static double dp2[1 << 13];
		static int masks[1 << 13];
		vi is;
		rep(s, 1 << N) if((s & (s-1)) != 0) {
			is.clear();
			rep(i, N) if(s >> i & 1)
				is.push_back(i);
			int n = is.size();
			rep(t, 1 << n) {
				int u = 0;
				rep(ix, n) if(t >> ix & 1)
					u |= 1 << is[ix];
				masks[t] = u;
			}
			double multgroupfailprob = 0;
			double multgroupsum = 0;
			rer(k, 0, maxv2) {
				rep(t, 1 << n)
					dp2[t] = 0;
				dp2[0] = 1;
				rep(ix, n) {
					int v = v_2[is[ix]];
					//1 / 2^{max(0, v_2[i] - k)}
					double prob1 = 1. / (1LL << max(0, v - k));
					double prob2 = k == 0 ? 0 : 1. / (1LL << max(0, v - (k - 1)));
					rep(t, 1 << ix) {
						double x = dp2[t];
						if(x < 1e-99) continue;
						dp2[t] = x * prob2;
						dp2[t | 1 << ix] = x * (prob1 - prob2);
					}
				}
				multgroupfailprob += dp2[(1 << n)-1];
				reu(t, 1, (1 << n)-1) {
					double x = dp2[t];
					if(x < 1e-99) continue;
					int u = masks[t];
					multgroupsum += x * (dp[u] + dp[s - u]);
				}
			}
			//dp2[t] = t が 0 になる確率
			rep(t, 1 << n)
				dp2[t] = 0;
			dp2[0] = 1;
			rep(ix, n) {
				long long p = P[is[ix]];
				double prob = 1. / p;
				rep(t, 1 << ix) {
					double x = dp2[t];
					if(x < 1e-99) continue;
					dp2[t] = x * (1 - prob);
					dp2[t | 1 << ix] = x * prob;
				}
			}
			double total = 0, failprob = 0;
			reu(t, 1, (1 << n)-1) {
				int u = masks[t];
				total += dp2[t] * (dp[u] + dp[s - u]);
			}
			failprob += dp2[(1 << n)-1];
			total += dp2[0] * multgroupsum;
			failprob += dp2[0] * multgroupfailprob;
			//E = 1 + failprob * E + total
			//E = (1 + total) / (1 - failprob)
			dp[s] = (1 + total) / (1 - failprob);
//			cerr << s << ": " << dp[s] << ", " << failprob << ", " <<total << endl;
		}
		double ans = dp[(1 << N)-1];
		printf("%.10f\n", ans);
	}
	return 0;
}