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.util.*;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
import java.io.*;
 
 
public final class Solution {
	
	static double PI = 3.1415926535;
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		PrintWriter writer = new PrintWriter(System.out);
		int t=sc.nextInt();
		while(t-->0) {
			int n=sc.nextInt();
			n=2*n;
			double a = PI*(n-2)/n, angle;
			angle=a;
			double len=0;
			while(angle>PI/2) {
				double x=PI-angle;
				len+=Math.cos(x);
				angle=a-x;
				//System.out.println(angle);
			}
			System.out.printf("%f", (1+2*len));
			System.out.println();
		}
		writer.flush();
		writer.close();
	}
}