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.*;
import java.util.*;
public class E87C1
{
    static PrintWriter out=new PrintWriter((System.out));
    public static void main(String args[])throws IOException
    {
        Reader sc=new Reader();
        int t=sc.nextInt();
        while(t-->0)
        {
            int n=sc.nextInt();
            solve(n);
        }
        out.close();
    }

    public static void solve(int n)
    {
        double ans=1/(Math.tan(Math.PI/(2*n)));
        out.println(ans);
    }

    static class Reader
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        public String next()
        {
            while(!st.hasMoreTokens())
            {
                try
                {
                    st=new StringTokenizer(br.readLine());
                }
                catch(Exception e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        public int nextInt()
        {
            return Integer.parseInt(next());
        }

        public long nextLong()
        {
            return Long.parseLong(next());
        }

        public double nextDouble()
        {
            return Double.parseDouble(next());
        }

        public String nextLine()
        {
            try
            {
                return br.readLine();
            }
            catch(Exception e)
            {
                e.printStackTrace();
            }
            return null;
        }

        public boolean hasNext()
        {
            String next=null;
            try
            {
                next=br.readLine();
            }
            catch(Exception e)
            {
            }
            if(next==null)
            {
                return false;
            }
            st=new StringTokenizer(next);
            return true;
        }
    }
}