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;
#pragma GCC optimize("-O3")
using ld = long double;
const long long mod = 1000000007;
const long long inf = 1000000000000000000;
const long long rk = 256;
const ld PI = 3.141592653589793;
ostream& operator<<(ostream& os, pair<long long, long long> const& x) {
  os << x.first << "," << x.second;
  return os;
}
template <class T>
ostream& operator<<(ostream& os, vector<T> const& x) {
  os << "{ ";
  for (auto& y : x) os << y << " ";
  return os << "}";
}
template <class T>
ostream& operator<<(ostream& os, set<T> const& x) {
  os << "{ ";
  for (auto& y : x) os << y << " ";
  return os << "}";
}
template <class Ch, class Tr, class Container>
basic_ostream<Ch, Tr>& operator<<(basic_ostream<Ch, Tr>& os,
                                  Container const& x) {
  os << "{ ";
  for (auto& y : x) os << y << " ";
  return os << "}";
}
template <class X, class Y>
ostream& operator<<(ostream& os, pair<X, Y> const& p) {
  return os << "[ " << p.first << ", " << p.second << "]";
}
const int mx = 1e6 + 50;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) {
    double n;
    cin >> n;
    double angle1 = PI / (4 * n);
    double angle2 = PI / (2 * n);
    double ans = cos(angle1) / sin(angle2);
    cout << setprecision(12) << ans << '\n';
  }
  return 0;
}