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;
bool prime(long long x) {
  if (x == 2) {
    return true;
  }
  if (x <= 1) {
    return true;
  }
  for (int i = 2; i <= sqrt(x); i++) {
    if (x % i == 0) {
      return false;
    }
  }
  return true;
}
void printa(long long n, long long a[]) {
  for (long long i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
  cout << endl;
}
void printv(vector<long long> v) {
  for (int i = 0; i < v.size(); i++) {
    cout << v[i] << " ";
  }
  cout << endl;
}
string Y = "YES\n";
string N = "NO\n";
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    cout << fixed << setprecision(20)
         << (1.0 / (2 * sin(double(3.1415926535897932384626) / (4 * n))))
         << endl;
  }
}