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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Comparator;

public class d {
    static BufferedReader s = new BufferedReader(new InputStreamReader(System.in));
    // static Scanner s=new Scanner(System.in);
    public static void main(String[] args) throws IOException {
        StringBuilder sb = new StringBuilder();
      String[] s1=s();
      int t=i(s1[0]);
      while(t-->0){
        String[] s2=s();
        int n=i(s2[0]);
         // System.out.println(Math.sin((0.0174533)*45));
          int i=(2*n-4)/4;
          double ans=0;
          double x=((double)2*n-2)* 180.0/(2.0*(double)n)-90;
          while(i-->0){
              ans+=Math.sin((0.0174533)*(x));
           //   System.out.println(x);
              x=((double)2* n-2)* 180.0/(2.0*(double)n)-(180.0-x);
          }
          System.out.println((double)1+2*ans);
      }
//        System.out.println(ans);
    }
    static int MAXN;
    static int[] spf;
    static void sieve() {
        spf[1] = 1;
        for (int i=2; i<MAXN; i++)
            spf[i] = i;
        for (int i=4; i<MAXN; i+=2)
            spf[i] = 2;
        for (int i=3; i*i<MAXN; i++)
        {
            if (spf[i] == i)
            {
                for (int j=i*i; j<MAXN; j+=i)

                    if (spf[j]==j)
                        spf[j] = i;
            }
        }
    }

    static String[] s() throws IOException {

        return s.readLine().trim().split("\\s+");
    }

    static int i(String ss) {
        return Integer.parseInt(ss);
    }
    static long l(String ss) {
        return Long.parseLong(ss);
    }
}

class Student12 {
    int  l;int r;
    public Student12(int l, int r) {
        this.l = l;
        this.r = r;
    }

    public String toString()
    {
        return this.l+" ";
    }
}
class Sortbyroll12 implements Comparator<Student12> {

    public int compare(Student12 a, Student12 b){
       /* if(b.r<a.r) return -1;
        else if(b.r==a.r) return 0;
        return 1;*/
        if(b.r-b.l>a.r-a.l) return 1;
        else if(b.r-b.l==a.r-a.l) {
            return a.l-b.l;
        }else return -1;/*
       if(b.l<a.l) return -1;
       else if(b.l==a.l) {
           return b.r-a.r;
       }
       return 1;*/
        //        return b.r- a.r;
        //        return (int) a.l-(int) b.l;
         /*   if(a.r<b.r) return -1;
            else if(a.r==b.r){
                if(a.r==b.r){
                    return 0;
                }
                if(a.r<b.r) return -1;
                return 1;}
            return 1;  */}
}