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#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;
}