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 findSubString(string str, string pat) {
  int len1 = str.length();
  int len2 = pat.length();
  map<char, int> hash_pat;
  map<char, int> hash_str;
  for (int i = 0; i < len2; i++) hash_pat[pat[i]]++;
  int start = 0, start_index = -1, min_len = INT_MAX;
  int count = 0;
  for (int j = 0; j < len1; j++) {
    hash_str[str[j]]++;
    if (hash_pat[str[j]] != 0 && hash_str[str[j]] <= hash_pat[str[j]]) count++;
    if (count == len2) {
      while (hash_str[str[start]] > hash_pat[str[start]] ||
             hash_pat[str[start]] == 0) {
        if (hash_str[str[start]] > hash_pat[str[start]]) hash_str[str[start]]--;
        start++;
      }
      int len_window = j - start + 1;
      if (min_len > len_window) {
        min_len = len_window;
        start_index = start;
      }
    }
  }
  if (start_index == -1) {
    return 0;
  }
  return min_len;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    double x;
    cin >> x;
    x = x * 2;
    double xDegrees = 180.0 / x;
    double result = tan((xDegrees * 3.141592653589793238) / 180);
    double ans = (1.0) / (2.0 * result);
    double a = ans * 2.0;
    cout << fixed << setprecision(7) << a << endl;
  }
  return 0;
}