Problem: The statement of this problem is the same as the statement of problem C2. The only difference is that, in problem C1, n is always even, and in C2, n is always odd.

You are given a regular polygon with 2 ⋅ n vertices (it's convex and has equal sides and equal angles) and all its sides have length 1. Let's name it as 2n-gon.

Your task is to find the square of the minimum size such that you can embed 2n-gon in the square. Embedding 2n-gon in the square means that you need to place 2n-gon in the square in such way that each point which lies inside or on a border of 2n-gon should also lie inside or on a border of the square.

You can rotate 2n-gon and/or the square.

Input

The first line contains a single integer T (1 ≤ T ≤ 200) — the number of test cases.

Next T lines contain descriptions of test cases — one per line. Each line contains single even integer n (2 ≤ n ≤ 200). Don't forget you need to embed 2n-gon, not an n-gon.

Output

Print T real numbers — one per test case. For each test case, print the minimum length of a side of the square 2n-gon can be embedded in. Your answer will be considered correct if its absolute or relative error doesn't exceed 10^{-6}.

Example

Input


3
2
4
200


Output


1.000000000
2.414213562
127.321336469

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void scanv(vector<T>& v, int n) {
  v.resize(n);
  for (int i = 0; i < n; i++) cin >> v[i];
}
template <typename T>
void scanvv(vector<vector<T> >& v, int n, int m) {
  v.resize(n);
  for (int i = 0; i < n; i++) scanv(v[i], m);
}
vector<pair<double, double> > p;
double dist(pair<double, double> a, pair<double, double> b) {
  return sqrt((a.first - b.first) * (a.first - b.first) +
              (a.second - b.second) * (a.second - b.second));
}
int main() {
  ios_base::sync_with_stdio(false);
  int cs, n, i, j;
  double theta, x, y, xx, yy, angle, d;
  cin >> cs;
  while (cs--) {
    cin >> n;
    n <<= 1;
    p.clear();
    p.resize(n + 1);
    theta = n * acos(-1.0) - 2 * acos(-1.0);
    theta /= n;
    ;
    d = .5 / cos(theta / 2);
    ;
    angle = (2 * acos(-1.0) / n) * (n / 2.0 - 1);
    d = d * sqrt(2 - 2 * cos(angle));
    ;
    printf("%.10lf\n", d);
  }
  return 0;
}