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;
template <typename T>
istream& operator>>(istream& in, vector<T>& v) {
  for (long long i = 0; i < ((int)(v).size()); i++) cin >> v[i];
  return in;
}
template <typename T, typename second>
istream& operator>>(istream& in, pair<T, second>& p) {
  cin >> p.first >> p.second;
  return in;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  for (auto& it : v) {
    os << it << " ";
  }
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& v) {
  for (auto& it : v) {
    os << it << " ";
  }
  return os;
}
template <typename T, typename second>
ostream& operator<<(ostream& os, const map<T, second>& v) {
  for (auto& it : v) {
    os << it << " ";
  }
  return os;
}
template <typename A, typename B>
ostream& operator<<(ostream& os, const pair<A, B>& p) {
  return os << '(' << p.first << "," << p.second << ')';
}
void solve() {
  long long n;
  cin >> n;
  cout << std::fixed << std::setprecision(15)
       << 1 / tan(3.14159265358979323846 / (2 * n)) << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}