#include <bits/stdc++.h>
using namespace std;
const int N = 110;
typedef long long LL;
LL f[N], sum;
// f(n)=(n-1)*(f(n-1)+f(n-2))
int main() {
    int m;
    scanf("%d", &m);

    while (m--) {
        int n;
        scanf("%d", &n);
        // base case
        sum = 1;
        f[1] = 0, f[2] = 1;

        //递推出错排次数
        for (int i = 3; i <= n; i++)
            f[i] = (i - 1) * (f[i - 1] + f[i - 2]);

        //总次数
        for (int i = 1; i <= n; i++) sum = sum * i;

        //比率
        printf("%.2lf%%\n", (double)f[n] / sum * 100);
    }
    return 0;
}