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.Scanner;

public class Main{

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scn =new Scanner(System.in);
	     int t=scn.nextInt();
	     while(t-->0){
	    	 int n=scn.nextInt();
	    	 double pi=3.14159265358979323846;
	    	 int s=2*n;
	    	    double ang=(double)(360)/s;
	    	    double ans=1;
	    	    for(double i=1;i<n/2;i++)
	    	    {
	    	        ans+=(double)(2*Math.sin(pi*ang*i/180));
	    	    }
	    	    System.out.println(ans);
	     }

	     
	}
	
	

}