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.*;
 
public class C {
 
	public static void main(String[] args)throws IOException {
		
		BufferedReader ob=new BufferedReader(new InputStreamReader(System.in));
		StringBuffer sb=new StringBuffer();
		
		int test=Integer.parseInt(ob.readLine());
		while(test-->0)
		{
			double n=Integer.parseInt(ob.readLine());
			n*=2;
			
			int a=((int)n-4)/4;
			double angle=(n-2)*180/n;
			double sum=1.0;
			
			double temp=angle-90;
			double pi=Math.PI/180;
			
			while(a-->0)
			{	
				double add=pi*temp;
				sum+=(2*Math.cos(add));
				
				double x=180.0-temp;
				temp=angle-x;
			}
			sb.append((float)sum+"\n");
			
		}
		
		System.out.println(sb);
	}
}