///by Saton.
#include<bits/stdc++.h>
#define PI acos(-1)
#define fi first
#define se second 
#define sz(a) ((int)a.size())
#define all(a) a.begin(), a.end()
#define int long long
#define ll __int128
#define DD double double
#define LD long double
#define rep(i,a,b) for(int i = (a);i <= (b);i ++)
#define lep(i,a,b) for(int i = (a);i >= (b);i --)
#define FLUSH fflush(stdout)
using namespace std;
typedef pair<int,int> PII;
const int N = 5e5 + 10,mod = 998244353;
int n,m,k;

const int M = 3e4;
int primses[M+5],cnt;
int cntl[M+5],val[M+5][20];
bool st[M+5];
void gar() {
    for(int i = 2;i <= M;i ++) {
        if(!st[i]) primses[cnt ++] = i;
        for(int j = 0;primses[j]*i <= M;j ++) {
            st[primses[j]*i] = true;
            if(i%primses[j]==0) break;
        }
    }
    for(int i = 0;i < cnt;i ++) {
        int len = 1;
        int tempv = primses[i];
        while(tempv <= M) {
            val[i][len ++] = tempv;
            tempv *= primses[i];
        }
        cntl[i] = len - 1; 
    }
}
double f[M+5];
void init() {
    fill(f,f+M+1,1);
    for(int i = 0;i < cnt;i ++) {
        for(int j = M;j >= 1;j --) {
            for(int k = 1;k <= cntl[i];k ++) {
                if(j>=val[i][k]) f[j] = max(f[j],f[j-val[i][k]]*(double)val[i][k]);
            }
        }
    }
}

void solve() {
    cin >> n;
    printf("%.9lf\n",log(f[n]));
} 

signed main() {
    gar();
    init();
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int T;
    cin >> T;
    while(T --) {
        solve();
    }
    // solve();
       
    return 0;
}  
/*   /\_/\
*   (= ._.)
*   / >  \>
*/