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;
double DEG_to_RAD(double d) { return d * acos(-1.0) / 180.0; }
auto main() -> int {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int tc;
  cin >> tc;
  cout << fixed << setprecision(8);
  while (tc--) {
    int n;
    cin >> n;
    double outangle = 180.0 * (2 * n - 2) / (2 * n);
    double inangle = 360.0 / (2 * n);
    double half_in_angle = 360.0 / (4 * n);
    double inside = 0.5 / sin(DEG_to_RAD(half_in_angle));
    double hi = inside * 2,
           lo = 0.5 / tan(DEG_to_RAD(half_in_angle)) * 2 - 0.01;
    while (lo + 1e-13 < hi) {
      double mid = (lo + hi) / 2;
      double shift = acos(mid / (2 * inside));
      double new_height = sin(DEG_to_RAD(outangle / 2) + shift) * 2 * inside;
      if (new_height <= mid)
        hi = mid;
      else
        lo = mid;
    }
    cout << hi << '\n';
  }
  return 0;
}