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 C2
{
    static final boolean RUN_TIMING = false;
    PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer tok;

    public void go() throws IOException
    {
        // in = new BufferedReader(new FileReader(new File("test.txt")));
        // out = new PrintWriter(new FileWriter(new File("output.txt")));
        tok = new StringTokenizer(in.readLine());
        int zzz = ipar();
        for (int zz = 0; zz < zzz; zz++)
        {
            ntok();
            int n = ipar();
            double angle = Math.PI / n / 2;
            double low = 0;
            double high = angle;
            while (high - low > 0.00000001) {
                double mid1 = low * 2/3 + high * 1/3;
                double mid2 = low * 1/3 + high * 2/3;
                double size1 = Math.max(mid1/angle / Math.sin(mid1), (angle-mid1)/angle / Math.sin(angle-mid1));
                double size2 = Math.max(mid2/angle / Math.sin(mid2), (angle-mid2)/angle / Math.sin(angle-mid2));
                if (size1 < size2) {
                    high = mid2;
                } else {
                    low = mid1;
                }
            }
            out.println(Math.max(low/angle / Math.sin(low), (angle-low)/angle / Math.sin(angle-low)));
        }

        out.flush();
        in.close();
    }

    public void ntok() throws IOException
    {
        tok = new StringTokenizer(in.readLine());
    }

    public int ipar()
    {
        return Integer.parseInt(tok.nextToken());
    }

    public int[] iapar(int n)
    {
        int[] arr = new int[n];
        for (int i = 0; i < n; i++)
        {
            arr[i] = ipar();
        }
        return arr;
    }

    public long lpar()
    {
        return Long.parseLong(tok.nextToken());
    }

    public long[] lapar(int n)
    {
        long[] arr = new long[n];
        for (int i = 0; i < n; i++)
        {
            arr[i] = lpar();
        }
        return arr;
    }

    public double dpar()
    {
        return Double.parseDouble(tok.nextToken());
    }

    public String spar()
    {
        return tok.nextToken();
    }

    public static void main(String[] args) throws IOException
    {
        long time = 0;
        time -= System.nanoTime();
        new C2().go();
        time += System.nanoTime();
        if (RUN_TIMING) {
            System.out.printf("%.3f ms%n", time / 1000000.0);
        }
    }
}