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.*;
	public class pract {
		 static int mysol(int n){
	    	 int arr[]=new int[n];
	    	 for(int i=0;i<n;i++)
	    		 arr[i]=i;
	    	 return n*2;
	     }
		 
	 public static void main(String[] args) {
		 Scanner scn = new Scanner(System.in);
		 int mT=scn.nextInt();
		 int mtcs=0;
		 for(int ll=0;ll<3;ll++){
			 int r=mysol(ll);
		 }
		 StringBuilder res=new StringBuilder();
		 while (mtcs++ < mT) {
				int mn = scn.nextInt();
				double x = Math.PI / (2 * mn);
				double ans = Math.sin(x);
				ans = 1 / ans;
				 for(int ll=0;ll<3;ll++){
					 int r=mysol(ll);
				 }
				double s = (Math.sqrt(ans * ans - 1.0));
				res.append(s + "\n");
			}
			System.out.print(res);
	 }
	 }