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 JAVA):

Solution:
import java.io.*;
import java.util.*;
import java.lang.*;
public class Main {
  public static void main(String[] args) throws IOException {
      BufferedReader b1 = new BufferedReader(new InputStreamReader(System.in));
      int t = Integer.parseInt(b1.readLine());
      while(t-->0) {
        //find length of longest diagonal using formula then apply pythagoras to find square side length
          int n = Integer.parseInt(b1.readLine());
          n *= 2;
          double d = (1)/Math.sin(Math.toRadians(180.0/n));
          double ans = Math.sqrt((d*d)-1);
          System.out.println(String.format("%.09f", ans));         
      }
  }
}