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.io.*;
public class cf645 {

	public static void main(String[] args) throws NumberFormatException, IOException {
	   BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	   int t=Integer.parseInt(br.readLine().trim());
	   while(t-->0) {
		  
//		   String str=br.readLine().trim();
//           char ch[]=str.toCharArray();
//           int c1=0,c2=0,c3=0,min=Integer.MAX_VALUE;
//           for(char c:ch) {
//        	   if(c=='1')c1++;
//        	   else if(c=='2')c2++;
//        	   else c3++;
//           }
//           if(c1==0 || c2==0||c3==0) {
//        	   System.out.println("0");
//           }else {
//        	    int b=0,c=0;
//        	   for(int i=0;i<str.length();i++){
//        		   if(ch[i]=='1') {
//        			   b=str.indexOf('2',i+1);
//        			   c=str.indexOf('3',i+1);
//        		   }else if(ch[i]=='2') {
//        			   b=str.indexOf('1',i+1);
//        			   c=str.indexOf('3',i+1);
//        		   }else {
//        			   b=str.indexOf('1',i+1);
//        			   c=str.indexOf('2',i+1);
//        		   }
//        		   if(b==-1 || c==-1)break;
//        		   int p=Math.max(b, c);
//        		   min=Math.min(min, p-i+1);
//        	   }
//        	   System.out.println(min);
//           }
		   
		   int n=Integer.parseInt(br.readLine().trim());
		   double r=Math.toRadians(90.0/n);
		   double tan=Math.tan(r);
		   double h=1/(tan);
		  // double ans=1/tan;
		   System.out.println(h);
		   
//		   double th=Math.toRadians(360.0/(4*n));
//           double tan=Math.tan(th);
//           double h=1/(2*tan);
	   }
	}
}