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 javax.management.QueryEval;
import  java.io.*;
import  java.math.*;
import java.nio.Buffer;
import java.sql.SQLSyntaxErrorException;
import  java.util.*;
import  java.text.*;
import java.util.regex.Pattern;
import java.util.stream.Collectors;


public class Main {

    public static long gcd(long a,long b){
        return b==0?a:gcd(b,a%b);
    }

    private static String acb;
    private static int func(int seed){
        return new Random(seed).nextInt();
    }
    public static double PI = Math.atan(1.0)*4;
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        int T=cin.nextInt();
        while(T-->0){
            double n=cin.nextInt();
            double ant=n;
            n=n*2.00;
            /*double d=((180*(n-2.00))/n)-90.00;
            System.out.println(d);
            double c=Math.sin(d);
            System.out.println(c);
            double long1=c*(ant-2);
            //System.out.println(long1);
            if(ant==2){
                System.out.println("1.000000000");
            }
            else{
                System.out.printf("%.9f\n",long1+1.00);
            }*/

            double ans=(1.0/(2.0*Math.sin(PI/n)))*Math.cos(360/n/2.0*(PI/180));
            System.out.printf("%.9f\n",ans*2.0);
                    //printf("%.9lf\n",ans*2);

        }

    }
}

class task implements Comparable<task>{

   public int x;
    public  task(int x){
        this.x=x;
    }
    @Override
    public int compareTo(task task) {
        return x-task.x;
    }
}