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;
string minWindow(string s, string t) {
  unordered_map<char, int> m, m1;
  int ele = 0;
  for (int i = 0; i < t.size(); i++) {
    m[t[i]]++;
    ele++;
  }
  int count = 0;
  int start = 0, window = INT_MAX, start_index = -1;
  for (int i = 0; i < s.size(); i++) {
    m1[s[i]]++;
    ele++;
    if (m1[s[i]] <= m[s[i]] && m[s[i]] != 0) count++;
    if (count == t.size()) {
      ele++;
      while (m1[s[start]] > m[s[start]] || m[s[start]] == 0) {
        if (m1[s[start]] > m[s[start]]) m1[s[start]]--;
        start++;
        ele++;
      }
      int win = i - start + 1;
      if (win < window) {
        window = win;
        ele--;
        start_index = start;
      }
    }
  }
  if (start_index == -1) return "";
  return s.substr(start_index, window);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long N;
    cin >> N;
    cout << fixed << setprecision(12)
         << 1.0 /
                tan(((360.0) / (2 * N * 2)) * (3.14159265358979323846 / 180.0))
         << endl;
  }
}