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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.text.DecimalFormat;
import java.util.Scanner;
import java.text.NumberFormat;
import java.math.RoundingMode;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC1 solver = new TaskC1();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskC1 {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int t = in.nextInt();
            DecimalFormat df = new DecimalFormat("#.########");
            df.setRoundingMode(RoundingMode.HALF_EVEN);
            for (int i = 0; i < t; i++) {
                int n = in.nextInt();
                int n_2 = 2 * n;
                if (n == 2) {
                    out.println(df.format(1 * 1.000000000));
                } else {
                    double angle = (Math.PI / (2.00000000000 * n));
                    String angleS = df.format(angle);
                    Double sss = Double.parseDouble(angleS);
                    double sinV = Math.tan(sss);
                    double ans = 1.00000000 / sinV;


                    /// area of polygon
                    // double area = (2*n) / (4.0 * Math.tan((Math.PI / 2*n)));
                    out.println(df.format(ans));
                }
            }
        }

    }
}