#include <iostream>
#include <algorithm>
#include <complex>
#include <utility>
#include <vector>
#include <string>
#include <queue>
#include <tuple>
#include <cmath>
#include <bitset>
#include <cctype>
#include <set>
#include <map>
#include <numeric>
#include <functional>
#define _overload3(_1,_2,_3,name,...) name
#define _rep(i,n) repi(i,0,n)
#define repi(i,a,b) for(ll i=ll(a);i<ll(b);++i)
#define rep(...) _overload3(__VA_ARGS__,repi,_rep,)(__VA_ARGS__)
#define all(x) (x).begin(),(x).end()
#define PRINT(V) cout << V << "\n"
#define SORT(V) sort((V).begin(),(V).end())
#define RSORT(V) sort((V).rbegin(), (V).rend())
using namespace std;
using ll = long long;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
inline void Yes(bool condition){ if(condition) PRINT("Yes"); else PRINT("No"); }
template<class itr> void cins(itr first,itr last){
    for (auto i = first;i != last;i++){
        cin >> (*i);
    }
}
template<class itr> void array_output(itr start,itr goal){
    string ans = "",k = " ";
    for (auto i = start;i != goal;i++) ans += to_string(*i)+k;
    if (!ans.empty()) ans.pop_back();
    PRINT(ans);
}
ll gcd(ll a, ll b) {
    return a ? gcd(b%a,a) : b;
}
const ll INF = 1e18;
const ll MOD = 1000000007;
typedef pair<ll,ll> P;
const ll MAX = 100005;
constexpr ll nx[8] = {1,0,-1,0,-1,-1,1,1};
constexpr ll ny[8] = {0,1,0,-1,-1,1,-1,1};
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int t;
    cin >> t;
    double a[] = {0,1.0000000000000000,1.0833333333333333,1.2569444444444444,1.5353009259259260,1.6915991512345676,2.0513639724794235};
    double p[6];
    p[0] = a[2]-1;
    p[1] = a[3]-a[2]*p[0]-1;
    p[2] = a[4]-a[3]*p[0]-a[2]*p[1]-1;
    p[3] = a[5]-a[4]*p[0]-a[3]*p[1]-a[2]*p[2]-1;
    p[4] = a[6]-a[5]*p[0]-a[4]*p[1]-a[3]*p[2]-a[2]*p[3]-1;
    p[5] = 1-p[0]-p[1]-p[2]-p[3]-p[4];
    rep(_,t){
        ll n;
        cin >> n;
        vector<double> e(1000010,0);
        for (ll i = n-1;i >= 0;i--){
            rep(k,1,7){
                e[i] += e[i+k]*p[k-1];
            }
            e[i] += 1.0;
        }
        printf("%.10f\n",e[0]);
    }
}