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;
int main() {
  int t;
  cin >> t;
  for (int q = 0; q < t; ++q) {
    int n;
    cin >> n;
    int k = n / 2 - 1;
    double pi = 3.14159265;
    cout << std::setprecision(10);
    double sum = 0.500000;
    double theta = pi * (n - 1) / n;
    for (int i = 1; i <= k; ++i) {
      sum += sin(i * (pi - theta));
    }
    sum = 2 * sum;
    double a = pi - theta;
    double d = pi - theta;
    double sum1 = sin(k * d / 2) / sin(d / 2) * sin((2 * a + (n - 1) * d) / 2);
    sum1 = (sum1 + 0.5) * 2;
    cout << sum << endl;
  }
}