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;
const double PI = atan(1) * 4;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long t, n;
  cin >> t;
  while (t-- != 0) {
    cin >> n;
    double ext = PI / n;
    double ht = 0;
    if (n % 2 == 0) {
      for (long long i = 0; i < n - 1; i++) {
        ht += 1.0 * sin(ext + ext * i);
      }
    } else {
      for (long long i = 0; i < (n - 1) / 2; i++) {
        ht += 1.0 * cos(ext + ext * i);
      }
      ht = 2 * ht + 1;
    }
    cout << fixed;
    cout << setprecision(7);
    cout << ht << "\n";
  }
  return 0;
}